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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();
}