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hdu1518 Square

2024-07-08 08:47:33   232人阅读

题目链接：

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

题目为：

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7839    Accepted Submission(s): 2526


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
 

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这个题目是搜索+剪枝优化..分析如下：

这个是给很多条棒子，然后看用某些棒子能拼成一个正方形。。
我的思路是如果搜索到4组，那么说明可以拼成一根木棒。。
剪枝的地方是：
1：如果有4组，那么久不要再搜索了。
2：为了避免重复，所以用了w这个变量。。。

代码为：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=20+10;
int a[maxn],vis[maxn];
int edge,average;
int t,sum,flag;
int m;
void dfs(int pos,int w,int count)
{
    if(pos==average)
    {
        count++;
        pos=0;
        w=1;
        if(count==4)
        {
            flag=1;
            return;
        }
    }
    if(flag)
        return;
    for(int i=w;i<=m;i++)
    {
       if(vis[i])
         continue;
       else
       {
           if(pos+a[i]<=average)
           {
               vis[i]=1;
               dfs(pos+a[i],i+1,count);
               vis[i]=0;
           }
       }
    }

}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        flag=0;
        sum=0;
        scanf("%d",&m);
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&a[j]);
            sum=sum+a[j];
        }
        average=sum/4;
        flag=0;
        if(sum%4!=0||m<4||a[m]>sum/4)
            printf("no\n");
        else
        {
            dfs(0,1,0);
            if(flag)
                printf("yes\n");
            else
                printf("no\n");
        }
    }
    return 0;
}


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