题干：
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 5Sample Output
yes
no
yes题目大意：
给定一些长度各异的木棒，请问是否可能把它们连接成一个正方形？解题报告：
这题是非常经典的dfs剪枝问题，但是一直弄不懂为什么必须要搞成 “一次dfs凑三条边 ”才可以，不能把dfs的功能限制成 凑一条边，然后进行三次dfs吗？（反正我是wa了。。。逃）
ps：与这道经典的剪枝题 相见恨晚啊！！
AC代码：
#include<bits/stdc++.h>using namespace std;
int n,sum;
int a[55];
bool vis[55];bool dfs(int cur, int st,int res) {if(cur == 4) return 1;if(res < 0) return 0 ;if(res == 0) {int bg = 1;while(vis[bg]) ++bg;for(int i = bg; i<=n; i++) {if(vis[i]) continue;vis[i]=1; if(dfs(cur+1,i+1,sum/4-a[i])) return 1;vis[i]=0;}return 0;}
// if(st > n) return 0;for(int i = st; i<=n; i++) {if(vis[i]) continue;vis[i]=1;if(dfs(cur,i+1,res - a[i])) return 1;vis[i]=0;}return 0;
}
bool cmp(const int & a,const int & b) {return a > b;
}
int main()
{int t;cin>>t;while(t--) {scanf("%d",&n);sum=0;int flag=1;for(int i = 1; i<=n; i++) scanf("%d",a+i),sum+=a[i]; if(sum%4 != 0 ){printf("no\n");continue;}memset(vis,0,sizeof vis);sort(a+1,a+n+1,cmp);if(dfs(1,1,sum/4)) printf("yes\n");else printf("no\n");}return 0 ;
}wa代码：
#include<bits/stdc++.h>using namespace std;
int n,sum;
int a[55];
bool vis[55];bool dfs(int st,int res) {if(res < 0) return 0 ;if(res == 0) return 1 ;for(int i = st; i<=n; i++) {if(vis[i]) continue;vis[i]=1;if(dfs(i+1,res - a[i])) return 1;vis[i]=0;}return 0;
}
bool cmp(const int & a,const int & b) {return a > b;
}
int main()
{int t;cin>>t;while(t--) {scanf("%d",&n);sum=0;int flag=1;for(int i = 1; i<=n; i++) scanf("%d",a+i),sum+=a[i]; if(sum%4 != 0 ){printf("no\n");continue;}memset(vis,0,sizeof vis);sort(a+1,a+n+1,cmp);int st = 1;for(int i = 1; i<=3; i++) {while(vis[st]) ++st;if(!dfs(st,sum/4)) {flag=0;break;} }if(flag) printf("yes\n");else printf("no\n");}return 0 ;
}ps：
求大神给个解释，，至今不知道为什么、、、
