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题目为：
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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：
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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：
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#include
#include
#include
#include
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 ); else { dfs(0,1,0); if(flag) printf(yes ); else printf(no ); } } return 0; }
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