uva 201 Squares
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; }