poj 1518 square dfs

最新推荐文章于 2019-10-14 21:27:43 发布

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本文探讨了一种算法问题——如何判断一系列不同长度的棍子能否拼接成一个正方形。通过对输入数据进行预处理和排序，利用深度优先搜索算法来尝试组合棍子形成正方形的边，并提供了一个C++实现示例。

Square
问题描述 :
Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?
输入:
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.
输出:

For each case, output a line containing “yes” if is is possible to form a square; otherwise output “no”.

样例输入:

3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5

样例输出:

yes
no
yes

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
const int maxn  = 100;
using namespace std;
int maxx = 0;
int a[maxn];
int vis[maxn];
bool flag ;
int avg;
int n;
void dfs(int num,int len,int start)
{
    if(num == 4)
    {
        flag =  true;
        return;
    }
    if(flag)
        return;
    if(len == avg )
    {
        dfs(num+1,0,0);
        if(flag)
            return;
    }
    for(int i = start; i < n;i++)
    {
        if(!vis[i]&&len+a[i]<=avg)
        {
            vis[i] =1;
            dfs(num,len+a[i],i+1);
            vis[i] = 0;
            if(flag)
                return;
        }
    }
}//成功的的变数，目前的长度，开始的位置
int main()
{
    int t;
    cin>>t;
    for(int  i = 0; i < t; i++)
    {
        int sum = 0;
        maxx = 0;
        cin>>n;
        for(int j = 0; j < n; j++)
        {
            cin>>a[j];
            if(a[j]>maxx)
                maxx = a[j];
            sum+=a[j];
        }
        avg= sum/4;
        if(maxx >avg||sum%4!=0)
        {
            cout<<"no"<<endl;
            continue;
        }
        sort(a,a+n);
        memset(vis,0,sizeof(vis));
        flag  = false;
        dfs(0,0,0);
        if(flag)
            cout<<"yes"<<endl;
        else cout<<"no"<<endl;
    }
    return 0;
}