POJ 2362 Square

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Square

Time Limit: 3000MS Memory Limit: 65536KTotal 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

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

Sample Output

yesnoyes

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;    }}

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