POJ 2362 Square

Language: Default

Square

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 21805		Accepted: 7616

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

这个题是一个DFS回溯的问题。我们可以求出来变长，我们只需要用已有的枝条凑出三个边长的长度，我们就可以返回成功。

我们还需要考虑到的是剪枝的简化运算。以下几个地方可以考虑去剪枝：1.首先所有木条的长度总和必须是4的倍数。2.满足1的前提下，我们可以计算正方形边长。如果最长的木条必须比边长要短。

为了简化运算，我们考虑将木条的长度排序，因为木条越长它的灵活性越差，我们需要尽早将长的木条凑进去。

贴上代码：

/*************************************************************************
	> File Name: Square.cpp
	> Author: Zhanghaoran0
	> Mail: chiluamnxi@gmail.com
	> Created Time: 2015年07月28日 星期二 16时59分48秒
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

using namespace std;

int T;
int n;
int a[30];
bool flag[30];
int sum = 0;

bool dfs(int num, int pos, int len){
    if(num == 3)
        return true;
    int i;
    for(i = pos; i >= 0; i --){
        if(!flag[i]){
            flag[i] = true;
            if(len + a[i] < sum){
                if(dfs(num, i - 1, len + a[i]))
                    return true;
            }
            else if(len + a[i] == sum)
                if(dfs(num + 1, n - 1, 0))
                    return true;
            flag[i] = false;
        }
    }
    return false;
}


int main(void){
    cin >> T;
    while(T --){
        cin >> n;
        bool temp = true;
        sum = 0;
        memset(flag, 0 ,sizeof(flag));
        if(n < 4){
             cout << "no"<< endl;
            continue;
        }
        for(int i = 0; i < n; i ++){
            cin >> a[i];
            sum += a[i];
        }
        if(sum % 4 != 0){
            cout << "no"<< endl;
            continue;
            }
        sum /= 4;
        sort(a, a + n);
        if(a[n - 1] > sum){
            cout << "no" << endl;
            continue;
        }
        if(dfs(0, n - 1, 0))
            cout << "yes" << endl;
    else
     cout <<"no"<< endl;
    }
}