hdu 1518(DFS+剪枝)

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13374    Accepted Submission(s): 4244

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

 

题意:问n根木棍能否组成一个正方形?

题解:排序+暴力+剪枝。。那些15ms的大神是怎么做到的。。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <stdlib.h>
using namespace std;

int v[25];
int n;
int sum = 0;
bool vis[25];
bool flag;
//cnt代表当前到第几根了,len代表当前木棍长度
void dfs(int cnt,int len,int now){
    if(cnt==4){
        flag = true;
        return;
    }
    for(int i=now;i<n;i++){
        if(flag) return;
        if(vis[i]||sum<len+v[i]) continue;
        if(sum==len+v[i]){
            vis[i] = true;
            dfs(cnt+1,0,0);
            vis[i] = false;
        }else if(sum>len+v[i]){
            vis[i] = true;
            dfs(cnt,len+v[i],i+1);
            vis[i] = false;
        }
    }
}
int cmp(int a,int b){
    return a>b;
}
int main(){
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d",&n);
        sum = 0;
        for(int i=0;i<n;i++){
            scanf("%d",&v[i]);
            sum+=v[i];
        }
        sort(v,v+n,cmp);
        if(sum%4!=0){
            printf("no
");
            continue;
        }
        sum/=4;
        if(v[0]>sum) {
            printf("no
");
            continue;
        }
        memset(vis,false,sizeof(vis));
        flag = false;
        dfs(0,0,0);
        if(flag) printf("yes
");
        else printf("no
");
    }
}