hdu 1518 Square
2014-07-26 07:53
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Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8260 Accepted Submission(s): 2687
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
<span style="font-size:18px;">#include<stdio.h> #include<iostream> #include<algorithm> using namespace std; #define M 25 int a[M],vis[M]; int n,m,sum; bool cmp(int x,int y) { return (x>y); } int dfs(int ans,int cur,int num) //ans表示当前位置 cur表示当前权值 num表示边数 { int i; if(num==4) return 1; for(i=ans;i<n;i++) { if(vis[i]) continue; if(cur+a[i]==m) // 满足边长 { vis[i]=1; if(dfs(0,0,num+1)) // 找下一条边 return 1; vis[i]=0; } else if(cur+a[i]<m) //不满足边长 { vis[i]=1; if(dfs(i,cur+a[i],num)) return 1; vis[i]=0; } } return 0; } int main () { int i,j,t; cin>>t; while(t--) { cin>>n; sum=0; for(i=0;i<n;i++) { cin>>a[i]; sum+=a[i]; } if(sum%4!=0) printf("no\n"); else { m=sum/4; // 边长 sort(a,a+n,cmp); if(m<a[0]) // 最大的数大于边长肯定不满足 printf("no\n"); else { memset(vis,0,sizeof(vis)); if(dfs(0,0,0)) printf("yes\n"); else printf("no\n"); } } } return 0; }</span>
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