1317: Square(DFS+剪枝)

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
• Raw

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

• Sample
  Output
• Raw

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;
}