题目大意:
有n行n列(2<=n<=9) 的小圆点,还有m条线段连接其中的一些黑点,统计这些线段连接成多少正方形的个数(每种边长分别统计);
解题思路:
模拟,用两个数组模拟行和列,统计时枚举定点扫描
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int H[10][10];
int V[10][10];
int main()
{
int n,m,x,y,T = 0;
char c;
while (~scanf("%d%d",&n,&m)) {
getchar();
for (int i = 1 ; i <= n ; ++ i)
for (int j = 1 ; j <= n ; ++ j)
V[i][j] = H[i][j] = 0;
for (int i = 1 ; i <= m ; ++ i) {
scanf("%c%d%d",&c,&x,&y);
getchar();
if (c == 'H')
H[x][y] = 1;
else
V[y][x] = 1;
}
if (T ++) printf("\n**********************************\n\n");
printf("Problem #%d\n\n",T);
int sum = 0;
for (int l = 1 ; l <= n ; ++ l) { //枚举正方形边长
int count = 0,flag = 0;
for (int i = 1 ; i+l <= n ; ++ i)
for (int j = 1 ; j+l <= n ; ++ j) { //枚举顶点
flag = 1;
for (int h = j ; h < j+l ; ++ h)
if (!H[i][h] || !H[i+l][h]) flag = 0;
for (int v = i ; v < i+l ; ++ v)
if (!V[v][j] || !V[v][j+l]) flag = 0;
count += flag;
}
sum += count;
if (count) printf("%d square (s) of size %d\n",count,l);
}
if (!sum) printf("No completed squares can be found.\n");
}
return 0;
}