Square

TimeLimit: 1 Second   MemoryLimit: 32 Megabyte

Totalsubmit: 1638   Accepted: 440

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

思路：从第一个数开始搜索，将其和与边长比对，相等则计数+1,计数达到3的时候说明可以组成，因为剩下那条必与边长相等，搜索过程注意剪枝，若某个数已被加入边长则不能重复计算，应将其标记，另外应在每一层递归时进行判断，看是否满足结束条件，以此来优化时间

#include<stdio.h>

#include<string.h>

int a[25],vis[25];

int con,temp,side,sum,flag,k;

　　//con用来记录边数，temp存放暂时的边长，用来与目标边长比对，index是每次查找的起始点（从上次结束的位置），非常重要，用此优化时间

void dfs(int con,int temp,int index)

{

int i;
if(3==con)
{
flag =1;
return;
}
if(temp==side)
{
dfs(con+1,0,0);
if(flag)
return;
}
for(i = index ;i < k;i++)
{
if(!vis[i])　　　　//判断此数是否已被用过
{
vis[i]=1;
dfs(con,temp+a[i],i+1);
if(flag)
return;
vis[i]=0;
}
}}int main(){
int n,m,max;
scanf("%d",&n);
while(n--)
{
k = sum =0;
flag = max =0;
memset(vis,0,sizeof(vis));
scanf("%d",&m);
while(m--)
{
scanf("%d",&a[k++]);
sum += a[k-1];
if(max<a[k-1])
max = a[k-1];
}
if(sum%4||max>sum/4)
{
printf("no\n");
continue;
}
side = sum/4;
dfs(0,0,0);
if(flag)
{
printf("yes\n");
continue;
}
printf("no\n");
}
return0;}