Square(TOJ 1398)

1398.   Square

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Time Limit: 2.5 Seconds   Memory Limit: 65536K
Total Runs: 685   Accepted Runs: 236

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 Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

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.

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

 

Output for Sample Input

yes
no
yes

Source: Waterloo Local Contest Sep. 21, 2002

 

代码：

 

#include<iostream> #include<algorithm> using namespace std; int len[25]; int visit[25]; int each; int n; bool b; int cmp(int a,int b) { return a>b; } void dfs(int i,int now,int nowlen) { if(now==3) { b=true; return ; } int j; if(b==false) { for(j=i;j<=n;j++) { if(!visit[j]&&nowlen+len[j]<=each) { visit[j]=1; if(nowlen+len[j]==each) dfs(1,now+1,0); else dfs(j+1,now,nowlen+len[j]); visit[j]=0; } } } } int main() { int t; cin>>t; while(t--) { cin>>n; int i; int sum=0; b=false; memset(visit,0,sizeof(visit)); for(i=1;i<=n;i++) { cin>>len[i]; sum+=len[i]; } sort(len+1,len+n+1,cmp); if(sum%4!=0) { cout<<"no"<<endl; continue; } else { each=sum/4; if(len[1]>each) { cout<<"no"<<endl; continue; } dfs(1,0,0);//dfs(int i,int now,int nowlen),i正在用哪根棍子,已经组装了now根,当前正在组装(还没完全)的棍子长度 if(b==true) cout<<"yes"<<endl; else cout<<"no"<<endl; } } return 0; }