dfs Square poj 2362

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本文探讨了使用不同长度的棍子构建正方形的问题，并提供了两种有效的算法解决方案。通过对输入数据的预处理和剪枝策略的应用，确保了算法的高效性和可行性。

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Square

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 20988		Accepted: 7314

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

Waterloo local 2002.09.21

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 50001
#include<algorithm>
using namespace std;
int a[maxn],vis[maxn],l,n,flag;
void dfs(int num,int len,int pos)
{
    if(num==4)
    {
        flag=1;
        return;
    }

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

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 5001
#include<algorithm>
using namespace std;
int a[maxn],vis[maxn],l,n;
int dfs(int num,int len,int pos)
{
    if(num==3)
        return 1;
    for(int i=pos;i>=0;i--)
    {
        if(!vis[i])
        {
            vis[i]=1;
            if(len+a[i]<l)
            {
               if(dfs(num,len+a[i],i-1))
                return 1;
            }
            else if(len+a[i]==l)
               if(dfs(num+1,0,n-1))
                  return 1;
                vis[i]=0;
        }
    }
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    int sum;
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        sum=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n);
        l=sum/4;
        if(n<4||sum%4!=0||l<a[n-1])
        {
            printf("no\n");
            continue;
        }
        if(dfs(0,0,n-1)) printf("yes\n");
        else printf("no\n");
    }
    return 0;
}