1317: Square(DFS+剪枝)

最新推荐文章于 2024-04-12 20:56:41 发布

王猛旗

最新推荐文章于 2024-04-12 20:56:41 发布

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分类专栏：搜索

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本文介绍了一种通过给定长度的棍子构建正方形的算法实现。算法首先判断输入的有效性，然后使用深度优先搜索（DFS）策略来尝试组合棍子形成正方形的边。文章提供了完整的C++代码示例，可用于理解如何通过编程解决问题。

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Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

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.

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

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

yes
no
yes

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
int data[250],state[250];
int ave,N,M,sum;
int dfs(int x,int pos,int len)
{
    int i;
    if(x==3)//如果有三条边都满足，结束
        return 1;
    for(i=pos;i>=0;i--){
        if(!state[i]){
            state[i]=1;
            if(len+data[i]<ave){
                if(dfs(x,i-1,len+data[i]))//i-1的作用是剪枝
                return 1;
            }
            if(len+data[i]==ave){
                if(dfs(x+1,M-1,0))
                return 1;
            }
            state[i]=0;
        }
    }
    return 0;
}
int main ()
{
    scanf("%d",&N);
    while(N--){
        sum=0;
        memset(state,0,sizeof(state));
        //memset(data,0,sizeof(data));
        scanf("%d",&M);
        for(int i=0;i<M;sum+=data[i],i++)
            scanf("%d",&data[i]);
        ave=sum/4;
        if(M<4||ave*4!=sum||ave<data[M-1]){
            printf("no\n");
            continue;
        }
        if(dfs(0,M-1,0))
            printf("yes\n");
        else
            printf("no\n");
        
    }
    return 0;
}

1317: Square(DFS+剪枝)

Description

Input

Output