hdu 1518 square (DFS)搜索

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Square

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14066 Accepted Submission(s): 4440

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

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

Sample Output

yesno
yes

//这题刚开始看的时候简单的以为直接暴力即可，求出sum 和然后sum/4 看能不能凑成一个正方形结果wrong

然后看数据1-20，并且时间度5000ms 应该是搜索算法了深度搜索

不过这个题也着实让我cao dan 调了一下午

/*DFS 搜索算法 */#include<iostream>#include<string.h>using namespace std;int a[35];bool vis[35];//true=1 false=0 bool flag; //  int n,sum;void dfs(int bian,int s_sum,int k){//cout<<bian<<"<<"<<"  "<<"-->>"<<s_sum<<"   "<<k<<endl;int i;if(bian==5)// 从1 开始 到n的话  则结束为 n+1{flag=true;return;}if(s_sum==sum)//如果  正好等于 那么 就凑下一条边  {dfs(bian+1,0,0);if(flag) return;} for( i=k;i<n;i++){//cout<<vis[i]<<endl;if(vis[i]==false&&s_sum+a[i]<=sum)//没有被使用 且  凑的这条边 小于平均边 {//cout<<"******"<<endl;vis[i]=true;//状态变化 dfs(bian,s_sum+a[i],i+1);//继续往下搜索 if(flag) return;vis[i]=false;//状态还原 }if(flag) return; } }int main(){int T,i;while(cin>>T){while(T--){memset(vis,0,sizeof(vis));//没进行一次 数组清零 int max=0; sum=0;// 记录边的总和 求平均边 cin>>n;for(i=0;i<n;i++){cin>>a[i];sum+=a[i];if(a[i]>max)//如果 里面 有 个边长度 大于 目标边 直接no max=a[i];} if(sum%4!=0||max>(sum/4))// 要求正方形边都是整数  平均边不是 直接 no {cout<<"no"<<endl;continue;}sum/=4;//求平均边 flag=false;//标志 dfs(1,0,0);if(flag)cout<<"yes"<<endl;elsecout<<"no"<<endl;}}}

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