poj 1518 square dfs

Square
问题描述 :
Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?
输入:
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.
输出:

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

样例输入:

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

样例输出:

yes
no
yes

#include <iostream>#include<stdio.h>#include<string.h>#include<algorithm>const int maxn  = 100;using namespace std;int maxx = 0;int a[maxn];int vis[maxn];bool flag ;int avg;int n;void dfs(int num,int len,int start){    if(num == 4)    {        flag =  true;        return;    }    if(flag)        return;    if(len == avg )    {        dfs(num+1,0,0);        if(flag)            return;    }    for(int i = start; i < n;i++)    {        if(!vis[i]&&len+a[i]<=avg)        {            vis[i] =1;            dfs(num,len+a[i],i+1);            vis[i] = 0;            if(flag)                return;        }    }}//成功的的变数，目前的长度，开始的位置int main(){    int t;    cin>>t;    for(int  i = 0; i < t; i++)    {        int sum = 0;        maxx = 0;        cin>>n;        for(int j = 0; j < n; j++)        {            cin>>a[j];            if(a[j]>maxx)                maxx = a[j];            sum+=a[j];        }        avg= sum/4;        if(maxx >avg||sum%4!=0)        {            cout<<"no"<<endl;            continue;        }        sort(a,a+n);        memset(vis,0,sizeof(vis));        flag  = false;        dfs(0,0,0);        if(flag)            cout<<"yes"<<endl;        else cout<<"no"<<endl;    }    return 0;}