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插头 DP code5-6

五、zju_3256
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#include<stdio.h>

#define N 7000 // 3^(n+1)
#define M 333  // all = 300+

int n,m;
int val[33],s[33],b[33],all,r[M],e[N];
int p[M][13],kk,H[M][13];

int A[M][M],B[M][M],tmp[M][M];

void mul(int A[M][M], int B[M][M])
{
	int i,j,k;
	long long w;
	for(i=0;i<all;i++)
	    for(j=0;j<all;j++)
	    {
	        w = 0;
	        for(k=0;k<all;k++) if(A[i][k] != 0 && B[k][j] != 0) w += (long long)A[i][k]*(long long)B[k][j];
	        if(w > 7777777) w = w-w/7777777*7777777;
	        tmp[i][j] = (int)(w);
	    }
	for(i=0;i<all;i++)
	    for(j=0;j<all;j++) A[i][j] = tmp[i][j];
}

int ok()
{
	int i,l,c[33];
	l = 0;
	for(i=1;i<=n+1;i++)
	{
	    b[i] = -1;
	    if(s[i] == 1) c[++l] = i;
	    else
	    if(s[i] == 2) {
	        if(l == 0) return 0;
	        b[c[l]] = i; b[i] = c[l]; l--;
	    }
	}
	if(l != 0) return 0;
	return 1;
}

void init()
{
	int i,j;
	val[1] = 1;
	for(i=2;i<=n+2;i++) val[i] = val[i-1]*3;

	for(i=0;i<=n+2;i++) s[i] = 0;

	all = 0;
	for(i=0;i<val[n+2];i++)
	{
	    e[i] = -1;
	    if(ok() == 1)
	    {
	        for(j=1;j<=n+1;j++) p[all][j] = b[j];
	        r[all] = i; e[i] = all;
	        all++;
	    }
	    s[1]++;
	    j = 1; while(s[j] > 2) { s[j] = 0; j++; s[j]++; }
	}
}

void abc()
{
	int i,j,k,LL,UU,sta;
	for(i=0;i<all;i++) for(j=0;j<all;j++) A[i][j] = (i==j)?1:0;

	for(k=0;k<all;k++) for(i=1;i<=n+1;i++) H[k][i] = r[k]/val[i]%3;

	for(i=1;i<=n;i++)
	{
	    for(j=0;j<all;j++) for(k=0;k<all;k++) B[j][k] = 0;

	    for(k=0;k<all;k++)
	    {
	        LL = H[k][i]; UU = H[k][i+1];

	        if(LL == 0 && UU == 0)
	        {
	            sta = r[k] + val[i] + val[i+1]*2;
	            if(e[sta] != -1) B[e[sta]][k] += 1;
	        }
	        else
	        if(LL == 0)
	        {
	            sta = r[k];
	            if(e[sta] != -1) B[e[sta]][k] += 1;

	            sta = r[k] + UU*(val[i]-val[i+1]);
	            if(e[sta] != -1) B[e[sta]][k] += 1;
	        }
	        else
	        if(UU == 0)
	        {
	            sta = r[k];
	            if(e[sta] != -1) B[e[sta]][k] += 1;

	            sta = r[k] + LL*(val[i+1]-val[i]);
	            if(e[sta] != -1) B[e[sta]][k] += 1;
	        }
	        else
	        if(LL == 2 && UU == 1)
	        {
	            sta = r[k] - val[i]*2 - val[i+1];
	            if(e[sta] != -1) B[e[sta]][k] += 1;
	        }
	        else
	        if(LL == 1 && UU == 1)
	        {
	            if(p[k][i+1] != -1)
	            {
	                sta = r[k]-val[i]-val[i+1]-val[p[k][i+1]];
	                if(e[sta] != -1) B[e[sta]][k] += 1;
	            }
	        }
	        else
	        if(LL == 2 && UU == 2)
	        {
	            if(p[k][i] != -1)
	            {
	                sta = r[k]-val[i]*2-val[i+1]*2+val[p[k][i]];
	                if(e[sta] != -1) B[e[sta]][k] += 1;
	            }
	        }
	    }

	    mul(B, A);
	    for(j=0;j<all;j++) for(k=0;k<all;k++) A[j][k] = B[j][k];
	}
	// change
	for(i=0;i<all;i++) for(j=0;j<all;j++)
	{
	    B[i][j] = 0;
	    if(e[r[i]/3] != -1)
	        B[i][j] = A[e[r[i]/3]][j];
	}

	int q[M],al=0;
	for(i=0;i<all;i++) if(r[i]%3 == 0) q[al++] = r[i];

	for(i=0;i<al;i++) for(j=0;j<al;j++)
	{
	    A[i][j] = B[e[q[i]]][e[q[j]]];
	}
	all = al;
	for(kk=0;;kk++) if(q[kk] == 3+val[n+1]*2) break;

}

void solve()
{
	int i,j;
	for(i=0;i<all;i++) for(j=0;j<all;j++) B[i][j]=(i==j)?1:0;
	while(m > 0)
	{
	    if(m%2 == 1) mul(B, A);
	    mul(A, A);
	    m /= 2;
	}

	if(B[kk][0] == 0)
	    printf("Impossible\n");
	else
	    printf("%d\n",B[kk][0]);
}

int main()
{
	while(scanf("%d %d",&n,&m) != EOF)
	{
	    init();
	    abc();
	    solve();
	}
	return 0;
}

六、hdu_1693
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#include<stdio.h>
#include<string.h>

#define N 5000 // 2^(m+1)

long long dp[2][N];
int H[N][15];

class DP {

public:
	int cas;
	DP() {
	    cas = 0;
	}

	int n,m,can[33][33];
	int val[33],LL,UU,sta,next;

	void input()
	{
	    int i,j;
	    scanf("%d %d",&n,&m);
	    for(i=1;i<=n;i++)
	        for(j=1;j<=m;j++) scanf("%d", &can[i][j]);
	}

	void abc()
	{
	    int i,j,k,u,y;

	    input();

	    u = 0;
	    for(i=0;i<=(1<<(m+1));i++) dp[u][i] = 0;
	    dp[u][0] = 1;

	    for(i=1;i<=n;i++)
	        for(j=1;j<=m;j++)
	        {
	            y = u; u = 1-u;
	            if(j == 1)
	            {
	                for(k=(1<<(m+1))-1;k>=0;k--)
	                {
	                    dp[y][k] = dp[y][k/2];
	                    if(k%2==1) dp[y][k] = 0;
	                }
	            }

	            for(k=0;k<(1<<(m+1));k++) dp[u][k] = 0;

	            for(k=0;k<(1<<(m+1));k++)
	            if(dp[y][k] > 0)
	            {
	                LL = H[k][j];
	                UU = H[k][j+1];

	                if(can[i][j] == 0)               // sta = 0
	                {
	                    if(LL == 0 && UU == 0)
	                    {
	                        dp[u][k] += dp[y][k];
	                    }
	                    continue;
	                }
	                                                 // sta = 1;
	                if(UU == 0 && LL == 0)
	                {
	                    sta = k+(1<<(j-1))+(1<<j);
	                    dp[u][sta] += dp[y][k];
	                }
	                else
	                if(LL == 0)
	                {
	                    sta = k;
	                    dp[u][sta] += dp[y][k];

	                    sta = k+((1<<(j-1)) - (1<<j));
	                    dp[u][sta] += dp[y][k];
	                }
	                else
	                if(UU == 0)
	                {
	                    sta = k;
	                    dp[u][sta] += dp[y][k];

	                    sta = k+(-(1<<(j-1))+(1<<j));
	                    dp[u][sta] += dp[y][k];
	                }
	                else
	                if(LL == 1 && UU == 1)
	                {
	                    sta = k-(1<<(j-1))-(1<<j);
	                    dp[u][sta] += dp[y][k];
	                }
	            }
	        }
	        cas++;
	        printf("Case %d: There are %lld ways to eat the trees.\n",cas,dp[u][0]);
	}

	void solve()
	{
	    int k,j,T;

	    for(k=0;k<(1<<(11+1));k++)
	        for(j=1;j<=11+1;j++)
	            H[k][j] = k/(1<<(j-1))%2;

	    scanf("%d",&T);
	    while(T-- > 0)
	        abc();
	}
};

int main()
{
	    DP dp;
	    dp.solve();
}

插头 DP code3-4

三、pku_1739
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#include<stdio.h>

class DP {

public:
	int n,m,can[33][33];
	int dp[200000],pre[200000],val[33],LL,UU,sta,next;
	int a[33],b[33],d[33],c[33];
	char p[200000][12];

	void input()
	{
	    int i,j;
	    char ch[13];

	    for(i=1;i<=n;i++)
	    {
	        scanf("%s",ch);
	        for(j=1;j<=m;j++) can[i][j] = ch[j-1]=='.'?1:0;
	    }
	}

	int ok()
	{
	    int i,l;
	    l = 0;
	    for(i=1;i<=m+1;i++)
	    {
	        b[i] = d[i] = -1;
	        if(a[i] == 1) c[++l] = i;
	        else
	        if(a[i] == 2)
	        {
	            if(l == 0) return 0;
	            b[c[l]] = i; d[i] = c[l];
	            l--;
	        }
	    }
	    if(l != 0) return 0;
	    return 1;
	}

	void init()
	{
	    int i,j,k;

	    val[1] = 1;
	    for(i=2;i<=m+2;i++) val[i] = val[i-1]*3;

	    for(i=0;i<=m+2;i++) a[i] = 0;
	    a[1] = -1;

	    for(i=0;i<=val[m+2];i++)
	    {
	        a[1]++;
	        k = 1; while(k <= m+1 && a[k]>2) { a[k]%=3; a[k+1]++; k++; }

	        if(ok() == 0) continue;

	        for(j=1;j<=m+1;j++) {
	            p[i][j] = -1;
	            if(b[j] != -1) p[i][j] = b[j];
	            if(d[j] != -1) p[i][j] = d[j];
	        }
	    }
	}

	void abc()
	{
	    int i,j,k;

	    input();
	    init();

	    for(i=0;i<=val[m+2];i++) dp[i] = 0;

	    dp[0] = 1;

	    for(i=1;i<=n;i++)
	        for(j=1;j<=m;j++)
	        {
	            if(j == 1)
	            {
	                for(k=val[m+2]-1;k>=0;k--)
	                {
	                    dp[k] = dp[k/3];
	                    if(k%3 != 0) dp[k] = 0;
	                }
	            }

	            for(k=0;k<val[m+2];k++)
	            {
	                pre[k] = dp[k];
	                dp[k] = 0;
	            }

	            for(k=0;k<val[m+2];k++)
	            if(pre[k] > 0)
	            {
	                LL = k/val[j]%3;
	                UU = k/val[j+1]%3;

	                if(can[i][j] == 0)
	                {
	                    if(LL == 0 && UU == 0)
	                    {
	                        dp[k] += pre[k];
	                    }
	                    continue;
	                }


	                if(UU == 0 && LL == 0)
	                {
	                    sta = k+val[j]+val[j+1]+val[j+1];

	                    dp[sta] += pre[k];
	                }
	                else
	                if(LL == 0)
	                {
	                    sta = k;
	                    dp[sta] += pre[k];

	                    sta = k+k/val[j+1]%3*(val[j]-val[j+1]);
	                    dp[sta] += pre[k];
	                }
	                else
	                if(UU == 0)
	                {
	                    sta = k;
	                    dp[sta] += pre[k];

	                    sta = k+k/val[j]%3*(-val[j]+val[j+1]);
	                    dp[sta] += pre[k];
	                }
	                else
	                if(LL == 2 && UU == 1)
	                {
	                    sta = k-val[j]-val[j]-val[j+1];
	                    dp[sta] += pre[k];
	                }
	                else
	                if(LL == 1 && UU == 1)
	                {
	                    if(p[k][j+1] > 0 && p[k][j+1] <= m+1)
	                    {
	                        sta = k-val[j]-val[j+1]-val[p[k][j+1]];
	                        dp[sta] += pre[k];
	                    }
	                }
	                else
	                if(LL == 2 && UU == 2)
	                {
	                    if(p[k][j] > 0)
	                    {
	                        sta = k-val[j]-val[j]-val[j+1]-val[j+1]+val[p[k][j]];
	                        dp[sta] += pre[k];
	                    }
	                }
	            }
	        }
	    printf("%d\n",dp[1+val[m]*2]);
	}

	void solve()
	{
	    while(true)
	    {
	        scanf("%d %d",&n,&m);
	        if(n == 0 && m == 0) break;
	        abc();
	    }
	}
};

int main() {
	    DP dp;
	    dp.solve();
	    return 0;
}

四、hdu_3377
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#include<stdio.h>

int n,m,can[33][33];
int dp[200000],pre[200000],val[33],LL,UU,sta,next;
int a[33],b[33],d[33],c[33],cas=0;
char p[200000][12];


class DP {

public:
	void input()
	{
	    int i,j;

	    for(i=1;i<=n;i++)
	        for(j=1;j<=m;j++) scanf("%d",&can[i][j]);
	}

	int ok()
	{
	    int i,l;
	    l = 0;
	    for(i=1+m;i>=1;i--)
	    {
	        b[i] = d[i] = -1;
	        if(a[i] == 2) c[++l] = i;
	        else
	        if(a[i] == 1)
	        {
	            if(l == 0) return 0;
	            d[c[l]] = i; b[i] = c[l];
	            l--;
	        }
	    }
	    if(l != 1) return 0;
	    return 1;
	}

	void init()
	{
	    int i,j,k;

	    for(i=0;i<=12;i++) a[i] = 0;
	    a[1] = -1;

	    for(i=0;i<=val[m+2];i++)
	    {
	        a[1]++;
	        k = 1; while(k <= m+1 && a[k]>2) { a[k]%=3; a[k+1]++; k++; }

	        if(ok() == 0) continue;

	        for(j=1;j<=m+1;j++) {
	            p[i][j] = -1;
	            if(b[j] != -1) p[i][j] = b[j];
	            if(d[j] != -1) p[i][j] = d[j];
	        }
	    }
	}

	void abc()
	{
	    int i,j,k,ans=-1000000000;

	    val[1] = 1;
	    for(i=2;i<=12;i++) val[i] = val[i-1]*3;

	    input();
	    init();

	    for(i=0;i<=val[m+2];i++) dp[i] = -1000000000;

	    //dp[0] = 0;

	    for(i=1;i<=n;i++)
	        for(j=1;j<=m;j++)
	        {
	            if(j == 1)
	            {
	                for(k=val[m+2]-1;k>=0;k--)
	                {
	                    dp[k] = dp[k/3];
	                    if(k%3 != 0) dp[k] = -1000000000;
	                }
	            }

	            for(k=0;k<val[m+2];k++)
	            {
	                pre[k] = dp[k];
	                dp[k] = -1000000000;
	            }

	            if(i == 1 && j == 1)
	            {
	                dp[2] = dp[6] = can[1][1];
	                continue;
	            }

	            if(i == n && j == m)
	            {
	                k = val[m]*2;
	                if(ans < pre[k]+can[n][m])
	                        ans = pre[k]+can[n][m];

	                k = val[m+1]*2;
	                if(ans < pre[k]+can[n][m])
	                        ans = pre[k]+can[n][m];
	                continue;
	            }

	            for(k=0;k<val[m+2];k++)
	            if(pre[k] > -1000000000)
	            {
	                LL = k/val[j]%3;
	                UU = k/val[j+1]%3;

	                if(UU == 0 && LL == 0)
	                {
	                    sta = k+val[j]+val[j+1]+val[j+1];

	                    if(pre[k] + can[i][j] > dp[sta])
	                        dp[sta] = pre[k] + can[i][j];

	                    if(dp[k] < pre[k]) dp[k] = pre[k];
	                }
	                else
	                if(LL == 0)
	                {
	                    sta = k;
	                    if(pre[k] + can[i][j] > dp[sta])
	                        dp[sta] = pre[k] + can[i][j];

	                    sta = k+k/val[j+1]%3*(val[j]-val[j+1]);
	                    if(pre[k] + can[i][j] > dp[sta])
	                        dp[sta] = pre[k] + can[i][j];
	                }
	                else
	                if(UU == 0)
	                {
	                    sta = k;
	                    if(pre[k] + can[i][j] > dp[sta])
	                        dp[sta] = pre[k] + can[i][j];

	                    sta = k+k/val[j]%3*(-val[j]+val[j+1]);
	                    if(pre[k] + can[i][j] > dp[sta])
	                        dp[sta] = pre[k] + can[i][j];
	                }
	                else
	                if(LL == 2 && UU == 1)
	                {
	                    sta = k-val[j]-val[j]-val[j+1];
	                    if(pre[k] + can[i][j] > dp[sta])
	                        dp[sta] = pre[k] + can[i][j];
	                }
	                else
	                if(LL == 1 && UU == 1)
	                {
	                    if(p[k][j+1] > 0 && p[k][j+1] <= m+1)
	                    {
	                        sta = k-val[j]-val[j+1]-val[p[k][j+1]];
	                        if(pre[k] + can[i][j] > dp[sta])
	                          dp[sta] = pre[k] + can[i][j];
	                    }
	                }
	                else
	                if(LL == 2 && UU == 2)
	                {
	                    if(p[k][j] > 0)
	                    {
	                        sta = k-val[j]-val[j]-val[j+1]-val[j+1]+val[p[k][j]];
	                        if(pre[k] + can[i][j] > dp[sta])
	                            dp[sta] = pre[k] + can[i][j];
	                    }
	                }
	            }
	        }
	    if(n == 1 && m == 1) ans = can[1][1];

	    printf("Case %d: %d\n",++cas,ans);
	}

	void solve()
	{
	    while(scanf("%d %d",&n,&m) != EOF)
	    {
	        abc();
	    }
	}
};

int main() {
	    DP dp;
	    dp.solve();
	    return 0;

}

插头 DP code1-2

一、hdu_1964
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#include<stdio.h>
#include<string.h>
#include<map>

int n,m,up[33][33],le[33][33];
int dp[200000],pre[200000],val[33],LL,UU,sta,next;
int a[33],b[33],d[33],c[33];
char p[200000][12];

int ok()
{
	int i,l;
	l = 0;
	for(i=1;i<=m+1;i++)
	{
	    b[i] = d[i] = -1;
	    if(a[i] == 1) c[++l] = i;
	    else
	    if(a[i] == 2)
	    {
	        if(l == 0) return 0;
	        b[c[l]] = i; d[i] = c[l];
	        l--;
	    }
	}
	if(l != 0) return 0;
	return 1;
}

void init()
{
	int i,j,k;

	val[1] = 1;
	for(i=2;i<=12;i++) val[i] = val[i-1]*3;

	for(i=0;i<=12;i++) a[i] = 0;
	a[1] = -1;

	for(i=0;i<=val[m+2];i++)
	{
	    a[1]++;
	    k = 1; while(k <= m+1 && a[k]>2) { a[k]%=3; a[k+1]++; k++; }

	    if(ok() == 0) continue;

	    for(j=1;j<=m+1;j++) {
	        p[i][j] = -1;
	        if(b[j] != -1) p[i][j] = b[j];
	        if(d[j] != -1) p[i][j] = d[j];
	    }
	}
}

void abc()
{
	int i,j,k,ans = 1000000;

	for(i=0;i<=val[m+2];i++) dp[i] = 1000000;

	dp[0] = 0;

	for(i=1;i<=n;i++)
	    for(j=1;j<=m;j++)
	    {
	        if(j == 1)
	        {
	            for(k=val[m+2]-1;k>=0;k--)
	            {
	                dp[k] = dp[k/3];
	                if(k%3 != 0) dp[k] = 1000000;
	            }
	        }

	        for(k=0;k<val[m+2];k++)
	        {
	            pre[k] = dp[k]; dp[k] = 1000000;
	        }

	        for(k=0;k<val[m+2];k++)
	        if(pre[k] < 1000000)
	        {
	            LL = k/val[j]%3;
	            UU = k/val[j+1]%3;

	            if(UU == 0 && LL == 0)
	            {
	                sta = k+val[j]+val[j+1]+val[j+1];
	                next = pre[k]+le[i][j+1]+up[i+1][j];
	                if(dp[sta] > next) dp[sta] = next;
	            }
	            else
	            if(LL == 0)
	            {
	                sta = k;
	                next = pre[k]+le[i][j+1];
	                if(dp[sta] > next) dp[sta] = next;

	                sta = k+k/val[j+1]%3*(val[j]-val[j+1]);
	                next = pre[k]+up[i+1][j];
	                if(dp[sta] > next) dp[sta] = next;
	            }
	            else
	            if(UU == 0)
	            {
	                sta = k;
	                next = pre[k]+up[i+1][j];
	                if(dp[sta] > next) dp[sta] = next;

	                sta = k+k/val[j]%3*(-val[j]+val[j+1]);
	                next = pre[k]+le[i][j+1];
	                if(dp[sta] > next) dp[sta] = next;
	            }
	            else
	            if(LL == 2 && UU == 1)
	            {
	                sta = k-val[j]-val[j]-val[j+1];
	                next = pre[k];
	                if(dp[sta] > next) dp[sta] = next;
	            }
	            else
	            if(LL == 1 && UU == 1)
	            {
	                if(p[k][j+1] > 0 && p[k][j+1] <= m+1)
	                {
	                    sta = k-val[j]-val[j+1]-val[p[k][j+1]];
	                    next = pre[k];
	                    if(dp[sta] > next) dp[sta] = next;
	                }
	            }
	            else
	            if(LL == 2 && UU == 2)
	            {
	                if(p[k][j] > 0)
	                {
	                    sta = k-val[j]-val[j]-val[j+1]-val[j+1]+val[p[k][j]];
	                    next = pre[k];
	                    if(dp[sta] > next) dp[sta] = next;
	                }
	            }
	            else
	            if(LL == 1 && UU == 2)
	            {
	                if(i == n && j == m)
	                {
	                    sta = k-val[j]-val[j+1]-val[j+1];
	                    next = pre[k];

	                    if(dp[sta] > next) dp[sta] = next;

	                    if(dp[sta] < ans) ans = dp[sta];
	                }
	            }

	        }
	    }

	printf("%d\n",ans);
}

int main()
{
	int i,j,T;
	char ch[1111];

	scanf("%d",&T);
	while(T--)
	{
	    scanf("%d %d",&n,&m);
	    gets(ch);
	    gets(ch);

	    for(i=0;i<=n+1;i++) for(j=0;j<=m+1;j++) up[i][j] = le[i][j] = 1000000;

	    for(i=1;i<=n;i++)
	    {
	        gets(ch);
	        for(j=2;j<=m;j++) le[i][j] = ch[j+j-2]-48;
	        gets(ch);

	        if(i<n)
	        {
	            for(j=1;j<=m;j++) up[i+1][j] = ch[j+j-1]-48;
	        }
	    }
	    init();
	    abc();
	}
	return 0;
}

二、timus_1519
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#include<stdio.h>

long long dp[60000],pre[60000];
int all,r[60000],e[1600000];
int p[60000][13];

class DP {

public:
	int n,m,can[33][33],last;
	int val[33],LL,UU,sta,next;
	int a[33],b[33],d[33],c[33];

	void input()
	{
	    int i,j;
	    char ch[15];
	    last = 0;
	    for(i=1;i<=n;i++)
	    {
	        scanf("%s",ch);
	        for(j=1;j<=m;j++)
	        {
	            can[i][j] = ch[j-1]=='.'?(++last):-1;
	        }
	    }
	}

	int ok()
	{
	    int i,l;
	    l = 0;
	    for(i=1;i<=m+1;i++)
	    {
	        b[i] = d[i] = -1;
	        if(a[i] == 1) c[++l] = i;
	        else
	        if(a[i] == 2)
	        {
	            if(l == 0) return 0;
	            b[c[l]] = i; d[i] = c[l];
	            l--;
	        }
	    }
	    if(l != 0) return 0;
	    return 1;
	}

	void init()
	{
	    int i,j,k;

	    val[1] = 1;
	    for(i=2;i<=m+2;i++) val[i] = val[i-1]*3;;

	    for(i=0;i<=m+2;i++) a[i] = 0;
	    a[1] = -1;

	    all = 0;

	    for(i=0;i<=val[m+2];i++)
	    {
	        a[1]++;
	        k = 1; while(k <= m+1 && a[k]>2) { a[k]%=3; a[k+1]++; k++; }

	        e[i] = -1;

	        if(ok() == 0) continue;

	        r[all] = i; e[i] = all;

	        for(j=1;j<=m+1;j++) {
	            p[all][j] = -1;
	            if(b[j] != -1) p[all][j] = b[j];
	            if(d[j] != -1) p[all][j] = d[j];
	        }

	        all++;
	    }
	}

	void abc()
	{
	    int i,j,k,l;

	    input();
	    init();

	    for(i=0;i<all;i++) dp[i] = 0;

	    dp[r[0]] = 1;

	    for(i=1;i<=n;i++)
	        for(j=1;j<=m;j++)
	        {
	            if(j == 1)
	            {
	                for(k=all-1;k>=0;k--)
	                {
	                    l = r[k]; l = l/3; l = e[l];

	                    if(l != -1)
	                    {
	                        dp[k] = dp[l];
	                        if(r[k]%3 != 0) dp[k] = 0;
	                    }
	                    else
	                        dp[k] = 0;
	                }
	            }

	            for(k=0;k<all;k++)
	            {
	                pre[k] = dp[k]; dp[k] = 0;
	            }

	            for(k=0;k<all;k++)
	            if(pre[k] > 0)
	            {
	                LL = r[k]/val[j]%3;
	                UU = r[k]/val[j+1]%3;

	                if(can[i][j] == -1)
	                {
	                    if(LL == 0 && UU == 0)
	                    {
	                        dp[k] += pre[k];
	                    }
	                    continue;
	                }

	                if(UU == 0 && LL == 0)
	                {
	                    sta = r[k]+val[j]+val[j+1]+val[j+1];

	                    dp[e[sta]] += pre[k];
	                }
	                else
	                if(LL == 0)
	                {
	                    sta = r[k];
	                    dp[e[sta]] += pre[k];

	                    sta = r[k]+r[k]/val[j+1]%3*(val[j]-val[j+1]);
	                    dp[e[sta]] += pre[k];
	                }
	                else
	                if(UU == 0)
	                {
	                    sta = r[k];
	                    dp[e[sta]] += pre[k];

	                    sta = r[k]+r[k]/val[j]%3*(-val[j]+val[j+1]);
	                    dp[e[sta]] += pre[k];
	                }
	                else
	                if(LL == 2 && UU == 1)
	                {
	                    sta = r[k]-val[j]-val[j]-val[j+1];
	                    dp[e[sta]] += pre[k];
	                }
	                else
	                if(LL == 1 && UU == 1)
	                {
	                    if(p[k][j+1] > 0 && p[k][j+1] <= m+1)
	                    {
	                        sta = r[k]-val[j]-val[j+1]-val[p[k][j+1]];
	                        dp[e[sta]] += pre[k];
	                    }
	                }
	                else
	                if(LL == 2 && UU == 2)
	                {
	                    if(p[k][j] > 0)
	                    {
	                        sta = r[k]-val[j]-val[j]-val[j+1]-val[j+1]+val[p[k][j]];
	                        dp[e[sta]] += pre[k];
	                    }
	                }
	                else
	                if(LL == 1 && UU == 2)
	                {
	                    if(can[i][j] == last)
	                    {
	                        sta = r[k]-val[j]-val[j+1]-val[j+1];
	                        dp[e[sta]] += pre[k];
	                    }
	                }
	            }
	        }
	    printf("%lld\n",dp[r[0]]);
	}

	void solve()
	{
	    while(scanf("%d %d",&n,&m) != EOF)
	    {
	        abc();
	    }
	}
};

int main() {
	    DP dp;
	    dp.solve();
	    return 0;
}

插头 DP

陈丹琦的大作 http://www.docin.com/p-46797997.html

http://acm.hdu.edu.cn/showproblem.php?pid=1964

http://acm.timus.ru/problem.aspx?space=1&num=1519

http://poj.org/problem?id=1739

http://acm.hdu.edu.cn/showproblem.php?pid=3377

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3256

http://acm.hdu.edu.cn/showproblem.php?pid=1693

http://acm.fzu.edu.cn/problem.php?pid=1977

http://poj.org/problem?id=3133

// timus_1519 有效状态间转移

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#include<stdio.h>

long long dp[60000],pre[60000];
int all,r[60000],e[1600000];
int p[60000][13];

class DP {

public:
	int n,m,can[33][33],last;
	int val[33],LL,UU,sta,next;
	int a[33],b[33],d[33],c[33];

	void input()
	{
		int i,j;
		char ch[15];
		last = 0;
		for(i=1;i<=n;i++)
		{
			scanf("%s",ch);
			for(j=1;j<=m;j++)
			{
				can[i][j] = ch[j-1]=='.'?(++last):-1;
			}
		}
	}

	int ok()
	{
		int i,l;
		l = 0;
		for(i=1;i<=m+1;i++)
		{
			b[i] = d[i] = -1;
			if(a[i] == 1) c[++l] = i;
			else
			if(a[i] == 2)
			{
				if(l == 0) return 0;
				b[c[l]] = i; d[i] = c[l];
				l--;
			}
		}
		if(l != 0) return 0;
		return 1;
	}

	void init()
	{
		int i,j,k;

		val[1] = 1;
		for(i=2;i<=m+2;i++) val[i] = val[i-1]*3;;

		for(i=0;i<=m+2;i++) a[i] = 0;
		a[1] = -1;

		all = 0;

		for(i=0;i<=val[m+2];i++)
		{
			a[1]++;
			k = 1; while(k <= m+1 && a[k]>2) { a[k]%=3; a[k+1]++; k++; }

			e[i] = -1;

			if(ok() == 0) continue;

			r[all] = i; e[i] = all;

			for(j=1;j<=m+1;j++) {
				p[all][j] = -1;
				if(b[j] != -1) p[all][j] = b[j];
				if(d[j] != -1) p[all][j] = d[j];
			}

			all++;
		}
	}

	void abc()
	{
		int i,j,k,l;

		input();
		init();

		for(i=0;i<all;i++) dp[i] = 0;

		dp[r[0]] = 1;

		for(i=1;i<=n;i++)
			for(j=1;j<=m;j++)
			{
				if(j == 1)
				{
					for(k=all-1;k>=0;k--)
					{
						l = r[k]; l = l/3; l = e[l];

						if(l != -1)
						{
							dp[k] = dp[l];
							if(r[k]%3 != 0) dp[k] = 0;
						}
						else
							dp[k] = 0;
					}
				}

				for(k=0;k<all;k++)
				{
					pre[k] = dp[k]; dp[k] = 0;
				}

				for(k=0;k<all;k++)
				if(pre[k] > 0)
				{
					LL = r[k]/val[j]%3;
					UU = r[k]/val[j+1]%3;

					if(can[i][j] == -1)
					{
						if(LL == 0 && UU == 0)
						{
							dp[k] += pre[k];
						}
						continue;
					}

					if(UU == 0 && LL == 0)
					{
						sta = r[k]+val[j]+val[j+1]+val[j+1];

						dp[e[sta]] += pre[k];
					}
					else
					if(LL == 0)
					{
						sta = r[k];
						dp[e[sta]] += pre[k];

						sta = r[k]+r[k]/val[j+1]%3*(val[j]-val[j+1]);
						dp[e[sta]] += pre[k];
					}
					else
					if(UU == 0)
					{
						sta = r[k];
						dp[e[sta]] += pre[k];

						sta = r[k]+r[k]/val[j]%3*(-val[j]+val[j+1]);
						dp[e[sta]] += pre[k];
					}
					else
					if(LL == 2 && UU == 1)
					{
						sta = r[k]-val[j]-val[j]-val[j+1];
						dp[e[sta]] += pre[k];
					}
					else
					if(LL == 1 && UU == 1)
					{
						if(p[k][j+1] > 0 && p[k][j+1] <= m+1)
						{
							sta = r[k]-val[j]-val[j+1]-val[p[k][j+1]];
							dp[e[sta]] += pre[k];
						}
					}
					else
					if(LL == 2 && UU == 2)
					{
						if(p[k][j] > 0)
						{
							sta = r[k]-val[j]-val[j]-val[j+1]-val[j+1]+val[p[k][j]];
							dp[e[sta]] += pre[k];
						}
					}
					else
					if(LL == 1 && UU == 2)
					{
						if(can[i][j] == last)
						{
							sta = r[k]-val[j]-val[j+1]-val[j+1];
							dp[e[sta]] += pre[k];
						}
					}
				}
			}
		printf("%lld\n",dp[r[0]]);
	}

	void solve()
	{
		while(scanf("%d %d",&n,&m) != EOF)
		{
			abc();
		}
	}
};

int main() {
	DP dp;
	dp.solve();
	return 0;
}

快速傅里叶变换计算大整数乘法 code

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// a->A, b->B C->c  用三次快速傅立叶变换。

#include <stdio.h>
#include <string.h>
#include <math.h>

#define N 50009

char s[N];
int La,Lb,a[N+N],b[N+N];

double pi = acos(-1.0);

struct Num {
	double a,b;
}
A[N+N],B[N+N],C[N+N];

Num operator+ (Num aa, Num bb) {
	Num ret;
	ret.a = aa.a+bb.a; ret.b = aa.b+bb.b;
	return ret;
}
Num operator- (Num aa, Num bb) {
	Num ret;
	ret.a = aa.a-bb.a; ret.b = aa.b-bb.b;
	return ret;
}
Num operator* (Num aa, Num bb) {
	Num ret;
	ret.a = aa.a*bb.a - aa.b*bb.b;
	ret.b = aa.a*bb.b + aa.b*bb.a;
	return ret;
}

Num W(int n, int k) {
	Num ret;
	ret.a = cos(-pi*k*2/n);
	ret.b = sin(-pi*k*2/n);
	return ret;
}

void DFT(int L, int R, Num from[], Num X[])
{
	if(L+1 == R)
	{
		X[L] = from[L];
		return;
	}

	int i,j,k;
	Num T;

	for(i=L;i<R;i++) X[i] = from[i];
	j = L; k = (L+R)/2;
	for(i=L;i<R;i+=2)
	{
		from[j++] = X[i]; from[k++] = X[i+1];
	}

	DFT(L, (L+R)/2, from, X);
	DFT((L+R)/2, R, from, X);

	for(i=L;i<(L+R)/2;i++)
	{
		T = X[i];
		X[i] = T + W(R-L, i-L)*X[i+(R-L)/2];
		X[i+(R-L)/2] = T - W(R-L, i-L)*X[i+(R-L)/2];
	}
}

int main()
{
	int i;
	while(scanf("%s",s) != EOF)
	{
		La = strlen(s);
		for(i=0;i<La;i++) a[i] = s[La-i-1]-48;
		scanf("%s",s);
		Lb = strlen(s);
		for(i=0;i<Lb;i++) b[i] = s[Lb-i-1]-48;

		i=1; while(i<La+Lb-1) i = i*2;
		for(;La<i;La++) a[La] = 0;
		for(;Lb<i;Lb++) b[Lb] = 0;

		for(i=0;i<La;i++) {
			A[i].a = a[i]; A[i].b = 0;
			B[i].a = b[i]; B[i].b = 0;
		}

		DFT(0, La, B, C);
		DFT(0, Lb, A, B);

		for(i=0;i<La;i++) B[i] = B[i]*C[i];
		DFT(0, La, B, C);

		C[La] = C[0]; b[0] = 0;
		for(i=1;i<=La;i++)
		b[i] = (int)(C[i].a/La + 0.5);

		for(i=La;i>0;i--)
		{
			b[i-1] += b[i]/10; b[i] %= 10;
		}

		i = 0; while(i < La && b[i] == 0) i++;
		for(;i<=La;i++) printf("%d",b[i]); printf("\n");
	}
	return 0;
}