squares id:201

最新推荐文章于 2019-09-01 15:41:43 发布

林伏案

最新推荐文章于 2019-09-01 15:41:43 发布

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本文链接：https://blog.youkuaiyun.com/qq_29169749/article/details/50945583

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本文介绍了一个用于游戏开发中计算特定大小正方形数量的算法实现过程，包括使用数组保存指令并进行暴力枚举的方法。通过实例演示如何在给定的二维数组中找出所有可能的正方形，并计算其数量。

A children’s board game consists of a square array of dots that contains lines connecting some of the pairs of adjacent dots. One part of the game requires that the players count the number of squares of certain sizes that are formed by these lines. For example, in the gure shown below, there are 3 squares
| 2 of size 1 and 1 of size 2. (The \size” of a square is the number of lines segments required to form a side.)
这里写图片描述
Your problem is to write a program that automates the process of counting all the possible squares

这里写图片描述

分析：
用两个数组来保存指令，然后对每个点进行暴力枚举。对每一个点就遍历一遍指令集。如果在指令集中有使得以该点作为左上顶点的正方形存在，则计数。

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

int V[10][10], H[10][10], cnt[10], n, m, kase = 0;
int a, b;
char cmd;

int is_exist(int size, int i, int j) {
    for(int h = j; h < j + size; h++)
        if(H[i][h] == 0 || H[i+size][h] == 0)   return 0;
    for(int v = i; v < i + size; v++)
        if(V[v][j] == 0 || V[v][j+size] == 0)   return 0;
    return 1;
}

int main() {

    while(~scanf("%d%d", &n, &m)) {
        memset(V, 0, sizeof(V));
        memset(H, 0, sizeof(H));
        memset(cnt, 0, sizeof(cnt));

        for(int i = 1; i <= m; i++) {
           cin >> cmd >> a >> b;
            if(cmd == 'H')      H[a][b] = 1;
            else                     V[b][a] = 1;
        }

        int size;
        for(size = 1; size < n; size++)
            for(int i = 1; i <= n - size; i++)
                for(int j = 1; j <= n - size; j++) {
                    if(is_exist(size, i, j))
                        cnt[size]++;
                }

        if(kase)       printf("\n**********************************\n\n");
        printf("Problem #%d\n\n", ++kase);
        int flag = 0;
        for(int i = 1; i < n; i++) {
            if(cnt[i] != 0) {
                flag = 1;
                printf("%d square (s) of size %d\n", cnt[i], i);
            }
        }
        if(flag == 0)   printf("No completed squares can be found.\n");
    }
}