ZOJ 2859 Matrix Searching 二维线段树 || 二维RMQ

题目：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1859

题意：

给定一个n*n的矩阵，有m个询问， 每个询问给出一个子矩阵的左上角和右下角坐标，问这个子矩阵内的最小值。

思路：

用二维线段树和二维RMQ都可以，没什么好说的。另外这个题数据少时限长，直接建n棵线段树都可以轻松水过，更暴力的方法也许也能过，大力出奇迹
二维线段树：

#include <bits/stdc++.h>using namespace std;const int N = 300 + 10, INF = 0x3f3f3f3f;int n;int minv[N<<2][N<<2];void push_upy(int ky, int kx){    minv[kx][ky] = min(minv[kx][ky<<1], minv[kx][ky<<1|1]);}void push_upx(int ky, int kx){    minv[kx][ky] = min(minv[kx<<1][ky], minv[kx<<1|1][ky]);}void buildy(int L, int R, int ky, int kx, int f){    if(L == R)    {        if(f) scanf("%d", &minv[kx][ky]);        else push_upx(ky, kx);        return;    }    int mid = (L + R) >> 1;    buildy(L, mid, ky << 1, kx, f);    buildy(mid + 1, R, ky << 1|1, kx, f);    push_upy(ky, kx);}void buildx(int L, int R, int kx){    if(L == R)    {        buildy(1, n, 1, kx, 1); return;    }    int mid = (L + R) >> 1;    buildx(L, mid, kx << 1);    buildx(mid + 1, R, kx << 1|1);    buildy(1, n, 1, kx, 0);}int queryy(int ly, int ry, int L, int R, int ky, int kx){    if(ly <= L && R <= ry) return minv[kx][ky];    int mid = (L + R) >> 1, ans = INF;    if(ly <= mid) ans = min(ans, queryy(ly, ry, L, mid, ky << 1, kx));    if(ry > mid) ans = min(ans, queryy(ly, ry, mid + 1, R, ky << 1|1, kx));    return ans;}int queryx(int lx, int rx, int ly, int ry, int L, int R, int kx){    if(lx <= L && R <= rx) return queryy(ly, ry, 1, n, 1, kx);    int mid = (L + R) >> 1, ans = INF;    if(lx <= mid) ans = min(ans, queryx(lx, rx, ly, ry, L, mid, kx << 1));    if(rx > mid) ans = min(ans, queryx(lx, rx, ly, ry, mid + 1, R, kx << 1|1));    return ans;}int main(){    int t, m;    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        buildx(1, n, 1);        scanf("%d", &m);        int x1, y1, x2, y2;        for(int i = 1; i <= m; i++)        {            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);            printf("%d\n", queryx(x1, x2, y1, y2, 1, n, 1));        }    }    return 0;}

二维RMQ：

#include <bits/stdc++.h>using namespace std;const int N = 300 + 5, INF = 0x3f3f3f3f;int a[N][N], dp[N][N][9][9];void ST(int n, int m){    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)            dp[i][j][0][0] = a[i][j];    for(int i = 0; (1<<i) <= n; i++)        for(int j = 0; (1<<j) <= m; j++)        {            if(i == 0 && j == 0) continue;            for(int row = 1; row <= n - (1<<i) + 1; row++)                for(int col = 1; col <= m - (1<<j) + 1; col++)                    if(i == 0) dp[row][col][i][j] = min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1]);                    else dp[row][col][i][j] = min(dp[row][col][i-1][j], dp[row+(1<<(i-1))][col][i-1][j]);        }}int RMQ(int x1, int y1, int x2, int y2){    int kx = log(x2 - x1 + 1.0) / log(2.0);    int ky = log(y2 - y1 + 1.0) / log(2.0);    int t1 = dp[x1][y1][kx][ky];    int t2 = dp[x2-(1<<kx)+1][y1][kx][ky];    int t3 = dp[x1][y2-(1<<ky)+1][kx][ky];    int t4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];    return min(min(t1, t2), min(t3, t4));}int main(){    int t, n, m;    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)                scanf("%d", &a[i][j]);        ST(n, n);        int x1, y1, x2, y2;        scanf("%d", &m);        while(m--)        {            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);            int ans = RMQ(x1, y1, x2, y2);            printf("%d\n", ans);        }    }    return 0;}