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最新推荐文章于 2021-08-29 18:25:25 发布

转载最新推荐文章于 2021-08-29 18:25:25 发布 · 115 阅读

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原文链接：http://www.cnblogs.com/zlrrrr/p/10113518.html

本文探讨了如何从一组不同长度的棍子中判断是否能拼接成一个正方形的问题，介绍了输入输出格式，提供了C++代码实现，并使用DFS算法进行解决。

https://vjudge.net/contest/67836#problem/I

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

3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5

Sample Output

yes
no
yes

代码：

#include <bits/stdc++.h>
using namespace std;

int n, ave, maxx, sum;
int a[50];
bool flag = false;
int vis[50];

void dfs(int num, int len, int start) {
    if(flag)
        return;

    if(num == 4) {
        flag = true;
        return ;
    }

    if(len == ave) {
        dfs(num + 1, 0, 1);
        if(flag)
            return ;
    }

    for(int i = start; i <= n; i ++) {
        if(vis[i] == 0 && len + a[i] <= ave) {
            vis[i] = 1;
            dfs(num, len + a[i], i + 1);
            vis[i] = 0;
            if(flag)
                return ;
        }
    }
}

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        scanf("%d", &n);
        maxx = 0, sum = 0;
        for(int i = 1; i <= n; i ++) {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        sort(a + 1, a + 1 + n);
        maxx = a[n];
        ave = sum / 4;
        if(sum % 4 != 0 || maxx > ave) {
            printf("no\n");
            continue;
        }
        memset(vis, 0, sizeof(vis));
        flag = false;
        dfs(0, 0, 1);
        if(flag)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}

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转载于:https://www.cnblogs.com/zlrrrr/p/10113518.html