HDU1518 Square（DFS）

                                              Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11151    Accepted Submission(s): 3588

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

 

题意： 给一列数， 问能否把这列数分成四组， 且每组数的和相同。

 

解法： 有DFS进行回溯。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int a[25], n, ans;
bool vis[25];

int dfs(int cur, int len, int pos)//cur表示用过的个数， len表示长度， pos表示位置 
{
    if(cur==n)
    return 1;
    for(int i=pos; i<n; i++)
    {
        if(vis[i]) continue; 
        if(len+a[i]<ans)
        {
            vis[i] = 1;
            if(dfs(cur+1, len+a[i], i+1)) return 1;
            vis[i] = 0;
            if(len==0) return 0;
            while(a[i]==a[i+1]&&i+1<n) ++i;//剪枝 
        }
        else if(len+a[i]==ans)
        {
            vis[i] = 1;
            if(dfs(cur+1, 0, 0)) return 1;
            vis[i] = 0;
            return 0;
        }
    }
    return 0;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int Sum = 0;
        scanf("%d", &n);
        for(int i=0; i<n; i++)
        {
            scanf("%d", &a[i]);
            Sum+=a[i];
        }
        if(Sum%4)
        {
            printf("no\n");
            continue;
        }
        ans = Sum/4;
        memset(vis, 0, sizeof(vis));
        int temp = dfs(0, 0, 0);
        printf("%s\n", temp?"yes":"no"); 
    }
    return 0;
}