这题求范围最小值，RMQ正好是用来解决这方面的。所以再适合只是了，又是离线静态输入输出的，所以时间比二维线段树快。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<map>

#include<queue>

#include<set>

#include<cmath>

#include<bitset>

#define mem(a,b) memset(a,b,sizeof(a))

#define lson i<<1,l,mid

#define rson i<<1|1,mid+1,r

#define INF 510010

#define maxn 310

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

int N;

int val[301][301];

int dp[maxn][maxn][9][9];

void RMQ_2D()

{

    for(int row = 1; row <= N; row++)

        for(int col = 1; col<=N; col++)

            dp[row][col][0][0] = val[row][col];

    int m = log(double(N)) / log(2.0);

    for(int i=0; i<=m; i++)

        for(int j=0; j<=m; j++)

        {

            if(i == 0 && j==0) continue;

            for(int row = 1; row+(1<<i)-1 <= N; row++)

                for(int col = 1; col+(1<<j)-1 <= N; 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]);  //水平划分

                    else dp[row][col][i][j] = min(dp[row][col][i-1][j] , dp[row+(1<<(i-1))][col][i-1][j]);    //竖直划分

                }

        }

}

int RMQ_2D(int x1,int x2,int y1,int y2)

{

    int kx = log(double(x2 - x1 +1)) / log(2.0);

    int ky = log(double(y2 - y1 +1)) / log(2.0);

    int m1 = dp[x1][y1][kx][ky];

    int m2 = dp[x2-(1<<kx)+1][y1][kx][ky];

    int m3 = dp[x1][y2-(1<<ky)+1][kx][ky];

    int m4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];

    return min( min(m1,m2), min(m3,m4) );

}

int main()

{

    int T;

    scanf("%d",&T);

    int M;

    int x1,y1,x2,y2;

    while(T--)

    {

        scanf("%d",&N);

        for(int i=1; i<=N; i++)

            for(int j=1; j<=N; j++)

                scanf("%d",&val[i][j]);

        RMQ_2D();

        scanf("%d",&M);

        while(M--)

        {

            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);

            printf("%d\n",RMQ_2D(x1,x2,y1,y2));

        }

    }

    return 0;

}