poj 2362 Square（dfs, 剪枝）

最新推荐文章于 2021-07-21 05:06:21 发布

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本文探讨了一种算法问题——利用不同长度的木棍构建正方形的可能性。通过输入一系列木棍长度，采用递归搜索和剪枝策略判断是否能形成正方形，并提供了完整的C++实现代码。

Square

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 23828		Accepted: 8253

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

Waterloo local 2002.09.21

思路：重在剪枝

代码：

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 25;
int a[maxn], n, avg, flag, have;
bool book[maxn];
void dfs(int fin, int num, int cur)
{
    if(flag) return ;
    if(fin == 4) { flag = 1; return ; }
    if(num == avg) { dfs(fin+1, 0, 0); return; }
    for(int i = cur; i < n; i++)
    {
        if(num+a[i] > avg) return ;<span style="white-space:pre">	</span>//剪枝2，因为对木棍排过序，搜到该棍过长，那后面的肯定更加长，不用搜了
        if(!book[i] && num+a[i] <= avg)
        {
            book[i] = 1;
            dfs(fin, num+a[i], i+1);
            book[i] = 0;
            if(flag) return ;<span style="white-space:pre">		</span>//剪枝3，一搜到就返回
        }
    }
}
int main(void)
{
    int t, maxlen;
    cin >> t;
    while(t--)
    {
        memset(book, 0, sizeof(book));
        avg = flag = maxlen = 0;
        scanf("%d", &n);
        for(int i = 0; i < n; i++) scanf("%d", &a[i]), avg += a[i], maxlen = max(maxlen, a[i]);
        if(avg % 4 || maxlen > avg/4) { printf("no\n"); continue ; }<span style="white-space:pre">	</span>//剪枝1，若最长的木棍大于边长，边长不能被4整除，肯定不行
        sort(a, a+n);<span style="white-space:pre">	</span>
        avg /= 4;
        dfs(0, 0, 0);
        if(flag) printf("yes\n");
        else printf("no\n");
    }
    return 0;
}