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Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12571 Accepted Submission(s): 4008
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int MAXN=25; int n,sum,l; int sticks[MAXN]; int vis[MAXN]; bool dfs(int dep,int len,int start,int num)//dep:dfs深度,len:目前sticks组成的长度,start:开始深搜的位置,num:组成的边数 { if(dep==n) { if(num==4) return true; else return false; } for(int i=start;i<n;i++) { if(!vis[i]&&len+sticks[i]<=l) { vis[i]=1; if(len+sticks[i]==l) { if(dfs(dep+1,0,0,num+1))//组成一个边从头开始搜 return true; } else { if(dfs(dep+1,len+sticks[i],i+1,num))//继续向前搜 return true; } vis[i]=0; } } return false; } int main() { int T; scanf("%d",&T); while(T--) { memset(vis,0,sizeof(vis)); sum=0; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&sticks[i]); sum+=sticks[i]; } if(sum%4!=0) { printf("no\n"); } else { sort(sticks,sticks+n);//排序 l=sum/4; if(dfs(0,0,0,0)) printf("yes\n"); else printf("no\n"); } } return 0; }
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原文地址:http://www.cnblogs.com/program-ccc/p/5626442.html