UVa 201- Squares

一开始确实没思路，如何存点和线段的问题困扰了很久，知道在网上看了前辈们的解法才开始用两个数组分别存横线和竖线。之后只要按照边数数正方形就好了。易错点是输出格式，要注意空行的输出。

题目链接：UVa 201

AC代码：

#include <iostream>
#include <cstring> 
using namespace std;
int H[10][10], V[10][10];
int main() {
	int n, m, cnt = 1, a, b;
	char ch;
	while (cin >> n) {
		cin >> m;
		memset(H, 0, sizeof(H));
		memset(V, 0, sizeof(V));
		while (m--) {
			cin >> ch >> a >> b;
			if (ch == 'H')
				H[a][b] = 1;
			else if (ch == 'V')
				V[b][a] = 1;
		}
		if (cnt > 1)
			cout <<endl<< "**********************************" << endl<<endl;
		cout << "Problem #" << cnt++ << endl<<endl;
		int sum = 0;
	for (int p = 1; p < n; p++) {  //按照边数由少至多数正方形  
		int count = 0;
		for(int i=1;i+p<=n;i++)
			for (int j = 1; j + p <= n; j++) {
				int flag = 1;
				for (int k = j; k < j + p; k++)
					if (!H[i][k] || !H[i + p][k])  //同时判断一对边 
						flag = 0;
				for (int l = i; l < i + p; l++)
					if (!V[l][j] || !V[l][j + p])
						flag = 0;
				count += flag;
			}
		sum += count;
		if (count)
			cout << count << " square (s) of size " << p << endl;
	}
	if (!sum)
		cout << "No completed squares can be found." << endl;
	}
	return 0;
}