Square

题目链接

Problem 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

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

题意

给你一些长度不等的棍子，是否能把它们组成一个正方形，（全部用上）

分析

用深搜把每种组合都遍历一遍，然后找出符合条件的，关键是dfs（），里面的该用什么参数不好想的。

dfs(int sidesNumber  ,int currentLength,int currentLocation)

sidesNumber:已经构成了几条边

currentLength: 当前正在构建边的长度

currentLocation： 正在构建边的位置

代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int stick[25];
int vis[25];
int op = 0, bianChang;
int m;
bool myCmp(int a, int b)
{
    return a > b;
}
void  dfs(int cur, int sum, int start)
{
    if (cur == 3 || op == 1)
    {
        op = 1;
        return ;
    }
    for (int i = start; i < m; i++)
    {
        if (vis[i] == 1)continue;
        if (sum + stick[i] == bianChang)//已经构建成一条边
        {  
            vis[i] = 1;
            dfs(cur + 1, 0, 0);//边的数目加一，准备构建下一条边，
                              //所以下一条边的长度初始化为零，还要从0位置开始构建，
            vis[i] = 0;
        }
        else if (sum + stick[i] < bianChang)//当前的边还没有构建完成，
        {
            vis[i] = 1;
            dfs(cur, sum + stick[i], i + 1);
            vis[i] = 0;
        }
    }
}
int main()
{
    int n;
    scanf("%d", &n);
    while (n--)
    {
        scanf("%d", &m);
        memset(stick, 0, sizeof(stick));
        memset(vis, 0, sizeof(vis));
        op = 0;
        int sum = 0;
        for (int i = 0; i < m; i++)
        {
            scanf("%d", &stick[i]);
            sum += stick[i];
        }
        if (sum % 4 != 0 || m < 4)
        {
            printf("no\n");
            continue;
        }
        sort(stick, stick + m, myCmp);
        bianChang = sum / 4;
        if (stick[0] > bianChang)
        {
            printf("no\n");
            continue;
        }
        dfs(0, 0, 0);//已经构成边的数目，正在构建某一条边的当前长度，从某个棍子开始构建。
        if (op == 1)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}
/*
3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5
*/