HDU 1518 Square（DFS）

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

Source

University of Waterloo Local Contest 2002.09.21

 #include<iostream>
 #include<cstring>
 using namespace std;
 int m,edge;//木棍的个数,木棍的边长
 int stick[];//木棍的长度
 bool vis[];//是否用过木棍
 bool flag;
 void dfs(int n,int len,int i)//n是第几条边，len是这条边已有长度,i从第几条边开始查找（防止超时）
 {
     if(n==)//终止条件
     {
         flag=true;
         return;
     }
     if(len==edge)
     {
         dfs(n+,,);
         if(flag==true)
             return ;
     }
     for(int j=i;j<=m;j++)
     {
         if(!vis[j])
         {
             if(stick[j]+len<=edge)
             {
                 vis[j]=true;
                 dfs(n,len+stick[j],j+);
                 if(flag==true)
                     return ;
                 vis[j]=false;
             }
         }
     }
 
 }
 int main()
 {
     int T;
     cin>>T;
     while(T--)
     {
         cin>>m;
         edge=;
         for(int i=;i<=m;i++)
         {
             cin>>stick[i];
             edge+=stick[i];
         }
         if(edge%!=)//第一个剪枝：是否能组成正方形
         {
             printf("no\n");
             continue;
         }
         edge/=;
         int i;
         for(i=;i<=m;i++)//第二个剪枝：木棍小于等于边长
         {
             if(stick[i]>edge)
                 break;
         }
         if(i!=m+)
         {
             printf("no\n");
             continue;
         }
         memset(vis,false,sizeof(vis));
         flag=false;
         dfs(,,);
         if(flag)
             printf("yes\n");
         else
             printf("no\n");
     }
     return ;
 }