squares id:201

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.)



Your problem is to write a program that automates the process of counting all the possible squares









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");
}
}