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划分树--查询区间k-th number

划分树 – 查询区间 k-th number

http://poj.org/problem?id=2104
http://acm.hdu.edu.cn/showproblem.php?pid=2665
http://acm.hdu.edu.cn/showproblem.php?pid=3727
http://acm.hdu.edu.cn/showproblem.php?pid=3473

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tr[ log(N) ][ N ]
1. 先对原来的数 稳定排序 ，tr[0][i] = 原先的数 a[i] 在排序后的位置。
2. dep = 0; s = 1; t = n;
3. 递归建树 
build_tree(dep, s, t)
{
    if(s >= t) return;
    mid = (s+t)/2;  j = s;  k = mid+1;
    for(i=s;i<=t;i++) if( tr[dep][i] <= mid )  tr[dep+1][j++] = tr[dep][i]; else tr[dep+1][k++] = tr[dep][i];
    //  把s 到 t 一分为二， s 到 t 的每个数 如果排序后它排在该区间的前半部分就移到下一层的前半部。
    //  如果要计算小于和大于 k-th number 的数的和要多算 sum[dep][x] 即 dep+1 层中 s 到 x(x<=mid) 的和 或 mid+1 到 x(x>mid) 的和。
    tr[dep][i] = j-1;  // s 到 t 区间， tr[dep][i] 记录 s 到 i 分到前半部分的最后位置
    build_tree(dep+1, s, mid);  build_tree(dep+1, mid+1, t);
}
4. 查找区间 [i,j] 中的 k-th number ，其中 1<=k<=j-i+1;
find_tree(dep, s, t, i, j, k)
{
    if(s == t) return s;
    v = i 到 j 中分到左边的数
    if(v >= k) return find_tree(dep+1, s, mid, ci, cj, k); // ci, cj 对应 i, j 分到前半部分的位置。 分到右半部分的和加到大于k-th number 上
    else    return find_tree(dep+1, mid+1, t, ci, cj, k-v); // 分到左半部分的和加到小于k-th number 上
 }
 
时间复杂度 O( n*log(n) 预处理， log(n) 查询 ) ，空间大小 n*log(n)
 
序列 ： 2 5 9 8 4 3 1
排序后  1 2 3 4 5 8 9
所以 原序列对应的最终位置为 2 5 7 6 4 3 1
 
        按最终位置分                                   指向分到前半部分的最后位置
tr[0][] = 2 5 7 6 4 3 1                        处理后 tr[0][] = 1 1 1 1 2 3 4
tr[1][] = 2 4 3 1 || 5 7 6                            tr[1][] = 1 1 1 2 || 5 5 6
tr[2][] = 2 1 || 4 3 || 5 6 || 7                      tr[2][] = 0 1 || 2 3 || 5 5 || 7
tr[3][] = 1 || 2 || 3 || 4 || 5 || 6

插头 DP code7-8

七、fzu_1977
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#include<stdio.h>

#define N 50000   // N = all
#define M 1600000  // M = 3^m

int n,m,can[33][33],last[33][33];

int all,val[33],r[N],e[M],p[N][13],b[33],w[33],cas=0;

long long dp[2][N];

int H[N][15];

void change()
{
	int i,j,tmp[33][33];
	for(i=1;i<=n;i++) for(j=1;j<=m;j++) tmp[m-j+1][i] = can[i][j];
	j = n; n = m; m = j;
	for(i=1;i<=n;i++)
	    for(j=1;j<=m;j++) can[i][j] = tmp[i][j];
}

void input()
{
	int i,j;
	char ch[33];

	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)
	{
	    scanf("%s",ch);
	    for(j=1;j<=m;j++)
	        if(ch[j-1] == 'O') can[i][j] = 1;
	        else
	        if(ch[j-1] == '*') can[i][j] = 2;
	        else
	            can[i][j] = 0;
	}

	if(n < m) change();

	int k=0;
	for(i=1;i<=n;i++) for(j=1;j<=m;j++)
	{
	    if(can[i][j] == 1) k++;
	    last[i][j] = k;
	}
}

int ok(int kk)
{
	int i,l,a[15],c[15];
	for(i=1;i<=m+1;i++) { a[i] = kk%3; kk /= 3; b[i] = -1; }
	l = 0;
	for(i=1;i<=m+1;i++)
	    if(a[i] == 1) c[++l] = i;
	    else
	    if(a[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<=m+2;i++) val[i] = val[i-1]*3;

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

	for(i=0;i<all;i++)
	    for(j=1;j<=m+1;j++)
	    {
	        H[i][j] = r[i]/val[j]%3;
	    }
}

void solve()
{
	int i,j,k,LL,UU,sta,u,y;

	long long ans = 0;

	u = 0;
	for(i=0;i<all;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=all-1;k>=0;k--)
	            {
	                if(e[r[k]/3] >= 0) dp[y][k] = dp[y][e[r[k]/3]];
	                if(r[k]%3 != 0) dp[y][k] = 0;
	            }
	        }

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

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

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

	            if(can[i][j] == 2)
	            {
	                if(LL == 0 && UU == 0)
	                    dp[u][k] += dp[y][k];
	            }

	            if(LL == 0 && UU == 0)
	            {
	                sta = r[k] + val[j] + val[j+1]*2;
	                dp[u][e[sta]] += dp[y][k];
	            }
	            else
	            if(LL == 0)
	            {
	                dp[u][k] += dp[y][k];

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

	                sta = r[k] - LL*val[j] + LL*val[j+1];
	                dp[u][e[sta]] += dp[y][k];
	            }
	            else
	            if(LL == 1 && UU == 1)
	            {
	                if(p[k][j+1] > 0)
	                {
	                    sta = r[k]-val[j]-val[j+1]-val[p[k][j+1]];
	                    dp[u][e[sta]] += dp[y][k];
	                }
	            }
	            else
	            if(LL == 2 && UU == 2)
	            {
	                if(p[k][j] > 0)
	                {
	                    sta = r[k]-val[j]*2-val[j+1]*2+val[p[k][j]];
	                    dp[u][e[sta]] += dp[y][k];
	                }
	            }
	            else
	            if(LL == 2 && UU == 1)
	            {
	                sta = r[k] - 2*val[j]-val[j+1];
	                dp[u][e[sta]] += dp[y][k];
	            }
	            else
	            if(LL == 1 && UU == 2)
	            {
	                if(r[k]-val[j]-val[j+1]*2 == 0 && last[i][j] == last[n][m])
	                {
	                    ans += dp[y][k];
	                }
	            }
	        }
	    }
	cas++;
	printf("Case %d: %lld\n",cas,ans);
}

int main()
{
	int i,T;
	scanf("%d",&T);

	m = 12;
	init(); r[all] = 1000000000;

	while(T-- > 0)
	{
	    input();
	    for(i=0;r[i]<val[m+2];i++); all = i;
	    solve();
	}
	return 0;
}
八、pku_3133
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#include<stdio.h>

#define N 60000+100 // 3^(m+1)

int n,m,can[33][33];

int dp[2][N],H[N][13],val[33];

void solve()
{
	int i,j,k,LL,UU,all,sta,u,y;

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

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

	        for(k=0;k<all;k++) dp[u][k] = 1000000;

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

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

	            if(can[i][j] == 0)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    if(dp[u][k] > dp[y][k]) dp[u][k] = dp[y][k];

	                    sta = k+val[j]+val[j+1];
	                    if(dp[u][sta] > dp[y][k] + 2) dp[u][sta] = dp[y][k]+2;

	                    sta = k+(val[j]+val[j+1])*2;
	                    if(dp[u][sta] > dp[y][k] + 2) dp[u][sta] = dp[y][k]+2;
	                }
	                else
	                if(LL == 0)
	                {
	                    if(dp[u][k] > dp[y][k]+1) dp[u][k] = dp[y][k]+1;

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

	                    sta = k-val[j]*LL+val[j+1]*LL;
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
	                }
	                else
	                if(LL == 1 && UU == 1)
	                {
	                    sta = k-val[j]-val[j+1];
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	                else
	                if(LL == 2 && UU == 2)
	                {
	                    sta = k-(val[j]+val[j+1])*2;
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	            }
	            else
	            if(can[i][j] == 2)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    sta = k+val[j];
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;

	                    sta = k+val[j+1];
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
	                }
	                else
	                if((LL == 1 && UU == 0) || (LL == 0 && UU == 1))
	                {
	                    sta = k-LL*val[j]-UU*val[j+1];
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	            }
	            else
	            if(can[i][j] == 3)
	            {
	                if(LL == 0 && UU == 0)
	                {
	                    sta = k+val[j]*2;
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;

	                    sta = k+val[j+1]*2;
	                    if(dp[u][sta] > dp[y][k] + 1) dp[u][sta] = dp[y][k]+1;
	                }
	                else
	                if((LL == 2 && UU == 0) || (LL == 0 && UU == 2))
	                {
	                    sta = k-LL*val[j]-UU*val[j+1];
	                    if(dp[u][sta] > dp[y][k]) dp[u][sta] = dp[y][k];
	                }
	            }
	        }
	    }
	if(dp[u][0] == 1000000) dp[u][0] = 0;
	printf("%d\n",dp[u][0]);
}

int main()
{
	int i,j;

	val[1] = 1;
	for(i=2;i<=9+2;i++) val[i] = val[i-1]*3;
	for(i=0;i<val[9+2];i++) for(j=1;j<=9+1;j++) H[i][j] = i/val[j]%3;

	while(scanf("%d %d",&n,&m) != EOF)
	{
	    if(n == 0 && m == 0)break;

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

	    solve();
	}
	return 0;
}

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();
}