HDU1518 DFS Square

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12231    Accepted Submission(s): 3897

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

 

Recommend

LL

解释都在代码里面，超时了好多次

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>


using namespace std;


int a[22];
bool vis[22];
bool flag;
int n,sum;


void DFS(int num,int l,int k)
{
    if(flag)
        return;
    if(num == 5)    ///四条边都已经找到
    {
        flag = 1;
        return;
    }
    if(l == sum)    ///找到一条边后，继续从0开始遍历
    {
        DFS(num + 1,0,0);
        if(flag)    ///减枝，节约时间
            return;
    }
    for(int i = k; i <= n; i ++)
        if(!vis[i] && a[i] + l <= sum)
        {
            vis[i] = 1;
            DFS(num,a[i] + l,i + 1);
            if(flag)    ///减枝，节约时间
                return;
            vis[i] = 0;
        }
}


int main()
{
    int t;
    scanf("%d",&t);
    while(t --)
    {
        sum = 0;
        scanf("%d",&n);
        for(int i = 1; i <= n; i ++)
            scanf("%d",&a[i]),sum+=a[i];
        if(sum % 4 != 0)    ///不能组成正方形
        {
            printf("no\n");
            continue;
        }
        sum /= 4;   ///每条边的长度
        int i;
        for( i = 1; i <= n; i ++)   ///是否有值超过每条边的长度
            if(a[i] > sum)
                break;
        if(i != n + 1)  ///有
        {
            printf("no\n");
            continue;
        }
        flag = 0;
        memset(vis,0,sizeof(vis));
        vis[0] = 1;
        DFS(1,0,0);///第一个是边，第二个是计算的每条边的长度，最后是当前所遍历的位置
        if(flag)
            printf("yes\n");
        else
            printf("no\n");
    }
}