题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1518
题目为：
Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7839 Accepted Submission(s): 2526
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
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这个题目是搜索+剪枝优化..分析如下：
这个是给很多条棒子，然后看用某些棒子能拼成一个正方形。。
我的思路是如果搜索到4组，那么说明可以拼成一根木棒。。
剪枝的地方是：
1：如果有4组，那么久不要再搜索了。
2：为了避免重复，所以用了w这个变量。。。
代码为：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=20+10;
int a[maxn],vis[maxn];
int edge,average;
int t,sum,flag;
int m;
void dfs(int pos,int w,int count)
{
if(pos==average)
{
count++;
pos=0;
w=1;
if(count==4)
{
flag=1;
return;
}
}
if(flag)
return;
for(int i=w;i<=m;i++)
{
if(vis[i])
continue;
else
{
if(pos+a[i]<=average)
{
vis[i]=1;
dfs(pos+a[i],i+1,count);
vis[i]=0;
}
}
}
}
int main()
{
scanf("%d",&t);
while(t--)
{
memset(vis,0,sizeof(vis));
flag=0;
sum=0;
scanf("%d",&m);
for(int j=1;j<=m;j++)
{
scanf("%d",&a[j]);
sum=sum+a[j];
}
average=sum/4;
flag=0;
if(sum%4!=0||m<4||a[m]>sum/4)
printf("no\n");
else
{
dfs(0,1,0);
if(flag)
printf("yes\n");
else
printf("no\n");
}
}
return 0;
}
hdu1518 Square