小学生玩ACM----深搜

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5771    Accepted Submission(s): 1836

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

#include <iostream>
#include <algorithm>
using namespace std;
int n,s,k,a[22],visit[22]={0};
void DFS(int sum,int x,int d)
{
    if(k||x==3)  //匹配好了三个边就行了，用k标记，并剪枝
    {
        k=1;return;
    }
    int i;
    for(i=d;i<n;i++)  //从d开始，起到剪枝作用，同时避免1-2-3和1-3-2这种重复
    {
        if(k)return;  //把k插入，剪枝
        if(!visit[i])
        if(sum+a[i]==s)  //加上这个长度后正好是一个边
        {
            visit[i]=1;
            DFS(0,x+1,0);
            visit[i]=0;
        }
        else if(sum+a[i]<s)  //加上这个长度后还不够形成一个边
        {
            visit[i]=1;
            DFS(sum+a[i],x,i+1);
            visit[i]=0;
        }
    }
}
bool cmp(int a,int b)
{
    return a<b;
}
int main(void)
{
    int m,i;
    cin>>m;
    while(m--&&cin>>n)
    {
        for(i=s=0;i<n;i++)
            cin>>a[i],s+=a[i];
        sort(a,a+n,cmp);
        if(s%4==0)
        {
            s/=4;
            k=0;
            DFS(0,0,0);  //里面的三个0分别是指当前合成的长度，当前已配好的边数，当前询问到的地点
            if(k)cout<<"yes"<<endl;
            else cout<<"no"<<endl;
        }
        else cout<<"no"<<endl;
    }
    return 0;
}

废话就不说了呢，这都还不会深搜那OrzOrz Orz Orz Orz OrzOrz Orz Orz Orz Orz