POJ 2445 Squares （未优化）

最新推荐文章于 2019-02-02 19:15:58 发布

sunny_pei

最新推荐文章于 2019-02-02 19:15:58 发布

阅读量433

点赞数

CC 4.0 BY-SA版权

分类专栏：算法模板

本文链接：https://blog.youkuaiyun.com/qq_19656301/article/details/82824068

算法模板专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个算法挑战，即编写程序来计算儿童棋盘游戏中由连接点形成的所有可能正方形的数量。输入包括一系列游戏板描述，每个描述包含n²个点和一些连接这些点的水平和垂直线段。任务是设计一个高效算法，遍历所有可能的正方形组合，并准确计数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A children’s board game consists of a square array of dots that contains lines connecting some dots. One part of the game requires that the players count the number of squares that are formed by these lines. For example, in the figure shown below, there are 3 squares, 2 of size 1 and 1 of size 2 , so the total number of squares is 3. (The “size” of a square is the number of lines segments required to form a side.)
在这里插入图片描述

Your task is to write a program to count the number of all the possible squares.

Input
The input represents a series of game boards. Each board consists of a description of a square array of n ^ 2 dots (where 2 <= n <= 1500) and some interconnecting horizontal and vertical lines. A record of a single board with n ^ 2 dots and m (m <= 300000 )interconnecting lines is formatted as follows:

Line 1: “n” the number of dots in a single row or column.
Line 2: “m” the number of interconnecting lines.
Each of the next m lines are one of following two types:

“H i j k” (1 <= i, j <= n, k >0, j + k <= n) indicates a horizontal line of length k from the dot in row i, column j to the dot in row i, column j + k.

“V i j k” (1 <= i, j <= n, k >0, i + k <= n) indicates a vertical line of length k from the dot in row i, column j to the dot in row i + k, column j.

The end of input is indicated by end-of-file.

Output
For each record print only one integer, which is the number of squares.

Sample Input
4
9
H 1 1 1
H 1 3 1
H 2 1 3
H 3 2 1
H 4 2 2
V 1 1 1
V 1 2 3
V 2 3 1
V 1 4 3
4
9
H 1 1 1
H 1 3 1
H 2 1 3
H 3 2 1
H 4 2 2
V 1 1 1
V 1 2 3
V 2 3 1
V 1 4 3

Sample Output
3
3

代码如下（这种方法数组需要开得非常大，超时，暂未优化）：

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;
 
int n,m;
int h[105][105];
int v[105][105];
int main()
{
    int cont=0;
    while(~scanf("%d %d",&n,&m))
    {
        char op;
        int a,b,c;
        memset(h,0,sizeof(h));
        memset(v,0,sizeof(v));
        while(m--)
        {
            cin>>op>>a>>b>>c;
            if(op=='H')
            	for(int i=b;i<b+c;i++)
                	h[a][i]=1;
            else
            	for(int j=a;j<a+c;j++)
                	v[j][b]=1;
        }
        int sum=0;
        for(int s=1;s<=n;s++)// 边从1-n
        {
            int ans=0,flag;// ans 统计 个数
            for(int i=1;i<=n-s;i++)// 暴力枚举
                for(int j=1;j<=n-s;j++)//  控制在n的范围内
                {
                    flag=1;
                    for(int k=j;k<j+s&&flag;k++)// 控制长度
                        if(!h[i][k]||!h[i+s][k]) flag=0;
                    for(int l=i;l<i+s&&flag;l++)
                        if(!v[l][j]||!v[l][j+s]) flag=0;
                    if(flag)
                        ans++;
                }
            sum+=ans;

        }

    	cout<<sum<<endl;
    }
 
    return 0;
}