ZOJ 2859 Matrix Searching

Matrix Searching

---

Time Limit: 10 Seconds      Memory Limit: 32768 KB

---

Given an n*n matrix A, whose entries Ai,j are integer numbers ( 1 <= i <= n, 1 <= j <= n ). An operation FIND the minimun number in a given ssub-matrix.

Input

The first line of the input contains a single integer T , the number of test cases.

For each test case, the first line contains one integer n (1 <= n <= 300), which is the sizes of the matrix, respectively. The next n lines with n integers each gives the elements of the matrix.

The next line contains a single integer N (1 <= N <= 1,000,000), the number of queries. The next N lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= n, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.

Output

For each test case, print N lines with one number on each line, the required minimum integer in the sub-matrix.

Sample Input

1
2
2 -1
2 3
2
1 1 2 2
1 1 2 1

Sample Output

-1
2

---

题意：T组数据，每组数据输入一个n*n的矩阵，再输入q个子矩阵，输出子每个矩阵中最小的元素。

每个矩阵建立一个结构体数组，将其对应的x，y坐标及其值val存入，按val的大小将结构体排序，对于每一子矩阵，从小到大搜索
找到满足在（lx，lr）到（rx，ry）范围的子矩阵中最小的元素便为所求的最小元素。

代码如下：

#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>using namespace std;struct maxi{    int x, y, val;} ma[90000+2];int cmp(const void *a, const void *b){    maxi *aa = (maxi*)a;    maxi *bb = (maxi*)b;    return aa->val - bb->val;}int main(){#ifdef test    freopen("in.txt", "r", stdin);#endif    int t, n, q, lx, ly, rx, ry;    scanf("%d", &t);    while(t--)    {        scanf("%d",&n);        int num = 0;        for(int i = 1; i <= n; i++)            for(int j = 1; j <= n; j++)            {                ma[num].x = i;                ma[num].y = j;                scanf("%d", &ma[num].val);                num++;            }        qsort(ma, num, sizeof(ma[0]), cmp); //按val值排序        scanf("%d", &q);        while(q--)        {            scanf("%d%d%d%d", &lx, &ly, &rx, &ry);            for(int i = 0; i < num; i++)                if((ma[i].x >= lx) && (ma[i].y >= ly) && (ma[i].x <= rx) && (ma[i].y <= ry))  //找在子矩阵中最小的元素                {                    printf("%d\n", ma[i].val);                    break;                }        }    }    return 0;}