rectangle（暴力求解）

Description:

Give you N rectangles.If you can pick exactly three pieces of those rectangles to form a larger rectangle?

input:

There are several testcases.

The first line is an integer T, indicating the number of testcases.

For each testcase:

The first line is a integer N and N is no more than 10.

The second line contains 2*N integers describing N rectangles.Each rectangle is described by 2 integers indicating as width and height.

All these integers in the second line are between [1,10000]

output

If you can do it, print Yes.Otherwise, print No instead.

sample input

2

4

1 1 1 1 1 2 2 2

4

1 1 2 2 10 10 20 20

sample output

Yes

No

#include<stdio.h>int a[110][2][2];int main() {    int T, n;    int i, j, k, x, y, z;    int flag;    scanf("%d", &T);    while (T--) {//不必使用for循环！        scanf("%d", &n);        for (i = 1; i <= n; i++) {            scanf("%d%d", &a[i][0][0], &a[i][0][1]);            a[i][1][0] = a[i][0][1];//0与1的取反，目的是方便后面的暴力求解            a[i][1][1] = a[i][0][0];        }        flag = 0;        for (i = 1; i <= n; i++) {            for (x = 0; x <= 1; x++)            for (j = 1; j <= n; j++) {                if (i == j) continue;                for (y = 0; y <= 1; y++)                for (k = 1; k <= n; k++) {                    if (i == k || j == k) continue;                    for (z = 0; z <= 1; z++) {                        if (a[i][x][0] + a[j][y][0] == a[k][z][0]                        && a[i][x][1] == a[j][y][1]) flag = 1;                        if (a[i][x][0] == a[j][y][0]                        && a[i][x][0] == a[k][z][0]) flag = 1;                    }                }            }        }//暴力求解！        if (flag) {            printf("Yes\n");        } else {            printf("No\n");        }    }}