Hdu 1518 Square

最新推荐文章于 2024-08-25 08:45:00 发布

颜洛滨

最新推荐文章于 2024-08-25 08:45:00 发布

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Square

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11643 Accepted Submission(s): 3737

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

Source

University of Waterloo Local Contest 2002.09.21

思路：简单的搜索题目，不过一开始没有思路，参考了网上的思路，从第一根木板开始搜索，如果符合条件，继续往下找，不符合，换另外一根，这里有一点点小的剪枝，就是最大的边不能大于目标的边，所有木板的长度和必须是四的倍数

AC代码如下：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn=20+5;
int a[maxn],vis[maxn];
int sum,n;
int ans;

void dfs(int cnt,int start,int tsum){

    if(cnt==4){
        ans=1;return ;
    }

    for(int i=start;i<n;i++){
        if(vis[i]) continue;
        if(a[i]+tsum>sum) continue;
        vis[i]=1;
        if(a[i]+tsum==sum) dfs(cnt+1,0,0);
        if(a[i]+tsum<sum) dfs(cnt,i+1,tsum+a[i]);
        vis[i]=0;
        if(ans)  return ;
    }
}
int main(){

    int t;
    scanf("%d",&t);
    while(t--){
       sum=n=0;
       int maxx=0;
       scanf("%d",&n);
       for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        sum+=a[i];
        if(a[i]>maxx) maxx=a[i];
       }
       if(sum%4){
        printf("no\n");
        continue;
       }
       sum/=4;
       memset(vis,0,sizeof(vis));
       if(sum<maxx){
        printf("no\n");
        continue;
       }
       ans=0;
       dfs(0,0,0);
       if(ans){
        printf("yes\n");
       }
       else printf("no\n");

    }
    return 0;
}