二维线段树之矩形树 _求解矩阵和问题

//zoj 2859
// 二维线段树之矩形树  求解矩阵和问题
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <deque>
#include <map>
#include <iostream>
using namespace std;
typedef long long LL;
const double pi = acos(-1.0);
const double e = exp(1);
//const int MAXN =2e5+10;
//const int N = 200010 * 4;
#define INT_MAX (1 << 30) - 1

const int maxn = 300 * 300 * 1.5;

struct node {
    short x1, x2, y1, y2;
    int min;
    int ch[4];
};
node tree[maxn];

int con[301][301];
int n, tol;

void build(int x1, int x2, int y1, int y2)
{
    int i;
    int k = ++tol;
    tree[k].x1 = x1, tree[k].x2 = x2;
    tree[k].y1 = y1, tree[k].y2 = y2;
    tree[k].min = 0;

    if(x1 == x2 && y1 == y2)
    {
        tree[k].min = con[x1][y1];
        memset(tree[k].ch, 0, sizeof(tree[k].ch));
        return ;
    }

    int midx = (x1 + x2) >> 1;
    int midy = (y1 + y2) >> 1;

    tree[k].ch[0] = tol + 1;
    build(x1, midx, y1, midy); //左下
    
    if(midx + 1 <= x2 && midy + 1 <= y2) //右上
    {
        tree[k].ch[1] = tol + 1;
        build(midx + 1, x2, midy + 1, y2);
    }
    else
    {
        tree[k].ch[1] = 0;
    }

    if(midy + 1 <= y2)  //左上
    {
        tree[k].ch[2] = tol + 1;
        build(x1, midx, midy + 1, y2);
    }
    else
    {
        tree[k].ch[2] = 0;
    }
    

    if(midx + 1 <= x2)  //右下
    {
        tree[k].ch[3] = tol + 1;
        build(midx + 1, x2, y1, midy);
    }
    else
    {
        tree[k].ch[3] = 0;
    }


    tree[k].min = tree[ tree[k].ch[0] ].min;

    for(i = 1; i < 4; i++)
    {
        if(tree[k].ch[i])
        {
            tree[k].min = min(tree[k].min, tree[ tree[k].ch[i] ].min);
        }
    }
    
}

bool judge(int k, int x1, int x2, int y1, int y2)
{
    if(x1 > tree[k].x2 || x2 < tree[k].x1 || y1 > tree[k].y2 || y2 < tree[k].y1)
        return false;
    return true;
}

int Query(int k, int x1, int x2, int y1, int y2)
{
    if(judge(k, x1, x2, y1, y2) == false)
        return INT_MAX;
    if(x1 <= tree[k].x1 && x2 >= tree[k].x2 && y1 <= tree[k].y1 && y2 >= tree[k].y2)
    {
        return tree[k].min;
    }

    int minx = Query(tree[k].ch[0], x1, x2, y1, y2);
    for(int i = 1; i < 4; i++)
    {
        minx = min(minx, Query(tree[k].ch[i], x1, x2, y1, y2));
    }
    return minx;
}



int main()
{
    int i, j, n, size, t;
    int x1, y1, x2, y2;
    scanf("%d", &t);
    while (t--)
    {
        //memset(T,0,sizeof(T));
        tol = 0;
        scanf("%d", &size);
        for (i = 1; i <= size; i++)
        {
            for (j = 1; j <= size; j++)
            {
                scanf("%d", &con[i][j]);
            }
        }
        build(1, size, 1, size);
        scanf("%d", &n);
        while (n--)
        {
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            printf("%d
",Query(1, x1, x2, y1, y2));
        }
    }

    return 0;
}