HDU 5319模拟

Description

Mr. Hdu is an painter, as we all know, painters need ideas to innovate , one day, he got stuck in rut and the ideas dry up, he took out a drawing board and began to draw casually. Imagine the board is a rectangle, consists of several square grids. He drew diagonally, so there are two kinds of draws, one is like ‘\’ , the other is like ‘/’. In each draw he choose arbitrary number of grids to draw. He always drew the first kind in red color, and drew the other kind in blue color, when a grid is drew by both red and blue, it becomes green. A grid will never be drew by the same color more than one time. Now give you the ultimate state of the board, can you calculate the minimum time of draws to reach this state.

 

Input

The first line is an integer T describe the number of test cases.
Each test case begins with an integer number n describe the number of rows of the drawing board.
Then n lines of string consist of ‘R’ ‘B’ ‘G’ and ‘.’ of the same length. ‘.’ means the grid has not been drawn.
1<=n<=50
The number of column of the rectangle is also less than 50.
Output
Output an integer as described in the problem description.

 

Output

Output an integer as described in the problem description.

 

Sample Input

2
4
RR.B
.RG.
.BRR
B..R
4
RRBB
RGGB
BGGR
BBRR 

 

Sample Output

3
6 

一个板， 不是矩形

两种颜色的刷子

R\         B/

一开始想了很久就是因为没看到这句

然后一遍就A了

求最少要画几刀

才能画成题目给的

我是从第一行开始 i，j

分R  B G判断它的斜下方的点能不能跟它连成一道

如果不行  就要画一笔  

不过行   就不加

#include <stdio.h>
#include <string.h>
#define N 55
char re[N][N]; 

int main(){
	int t, n;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		int count = 0;
		memset(re, 0, sizeof(re));
		for (int i = 0; i < n; i++)
			scanf("%s", &re[i]);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < strlen(re[i]); j++) {
				if ((re[i][j] == 'R' || re[i][j] == 'G') && !(re[i + 1][j + 1] == 'R' || re[i + 1][j + 1] == 'G'))
					count++;
				if ((re[i][j] == 'B' || re[i][j] == 'G') && !(re[i + 1][j - 1] == 'B' || re[i + 1][j - 1] == 'G'))
					count++;
				}
			}
		printf("%d\n", count);
	}
	return 0;
}