POJ2362 Square(dfs)

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Description

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

Input

The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

Output

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

Sample Input

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

Sample Output

yesnoyes

题意：

问给出的棍子可不可以凑成一个正方形

Note:

三处剪纸: 1.因为是正方形，棍子之和需整除4.

2.最短的木棒要比棍子平局数大。

3.dfs判断条件时只要三根棍子拼好即可。

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int maxn = 30;int n, side, stick[maxn];bool vis[maxn];bool cmp(int a, int b){return a > b;}bool dfs(int num, int len, int s){if(num == 3) return true;for(int i = s; i < n; ++i) {if(vis[i]) continue;vis[i] = true;if(len + stick[i] < side) {if(dfs(num, len + stick[i], i)) return true;}else if(len + stick[i] == side) {if(dfs(num + 1, 0, 0)) return true;}vis[i] = false;}return false;}int main(int argc, char const *argv[]){int t;scanf("%d", &t);while(t--) {memset(stick, 0, sizeof(stick));memset(vis, false, sizeof(vis));int sum = 0;scanf("%d", &n);for(int i = 0; i < n; ++i) {scanf("%d", &stick[i]);sum += stick[i];}if(n < 4 || sum % 4 != 0) {printf("no\n");continue;}sort(stick, stick + n, cmp);side = sum / 4;if(side < stick[0]) {printf("no\n");continue;}if(dfs(0, 0, 0)) printf("yes\n");else printf("no\n");}return 0;}

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