HDU1518 & POJ2362 & ZOJ1909 Square(DFS,剪枝是关键呀)

Square

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 4

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

yesnoyes

 

Source

University of Waterloo Local Contest 2002.09.21

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1518

题意:根据已知边的长度,问能否构成一个正方形.

解决超时是关键!

AC代码:

#include <iostream>#include <cstring>#include <algorithm>using namespace std;int a[25];int n,averlen;bool flag;bool vis[25];void DFS(int num,int len,int start)//成功边数,目前长度,开始位置{    if(flag)        return ;    if(num==4)    {        flag=true;        return;    }    if(len==averlen)    {        DFS(num+1,0,0);        if(flag)            return ;    }    for(int i=start;i<n;i++)    {        if(!vis[i]&&len+a[i]<=averlen)        {            vis[i]=true;            DFS(num,len+a[i],i+1);            vis[i]=false;            if(flag)                return ;        }    }}int main(){    int t;    cin>>t;    while(t--)    {        cin>>n;        int sum=0,maxlen=0;        for(int i=0;i<n;i++)        {            cin>>a[i];            sum+=a[i];            if(a[i]>maxlen)                maxlen=a[i];        }        averlen=sum/4;        if(sum%4!=0||maxlen>averlen)        {            cout<<"no"<<endl;            continue;        }        sort(a,a+n);        memset(vis,false,sizeof(vis));        flag=false;        DFS(0,0,0);        if(flag)            cout<<"yes"<<endl;        else            cout<<"no"<<endl;    }    return 0;}