hdoj 1518 Square 【dfs+剪枝】

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9930    Accepted Submission(s): 3254

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
   
   
   
   

 

dfs+剪枝。。。

 

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
bool cmp(int x,int y)
{
    return x>y;
}
int aver;
int stick[21];
int visit[21];
int n;
int exist;//判断是否匹配成功 
void dfs(int have,int num,int pos)//分别记录已匹配长度 已经匹配组数 当前搜索序号
{
    int i,j;
    if(num==4) 
    {
        exist=1;
        return ;
    }
    for(i=pos;i<n;i++)
    {
        if(i&&!visit[i-1]&&stick[i]==stick[i-1])//剪枝 
        continue;
        if(visit[i])
        continue;
        if(have+stick[i]==aver)
        {
            visit[i]=1;
            dfs(0,num+1,0);
            visit[i]=0;
        }
        else if(have+stick[i]<aver)
        {
            visit[i]=1;
            dfs(have+stick[i],num,i+1);
            visit[i]=0;
        }
        if(have==0)//节约时间的一步,若当前匹配失败 即使已经匹配成3对 也要跳出循环 
        break;//没有这句话跑800多ms 有的话只用78ms 
    } 
} 
int main()
{
    int t;
    int sum;
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&stick[i]);
            sum+=stick[i];
        }
        if(sum%4!=0)
        {
            printf("no\n");
            continue;
        }
        aver=sum/4;
        sort(stick,stick+n,cmp);
        if(stick[0]>aver)
        {
            printf("no\n");
            continue;
        }
        memset(visit,0,sizeof(visit));
        exist=0;
        dfs(0,0,0);
        if(exist)
        printf("yes\n");
        else
        printf("no\n");
    }
    return 0;
}