Painter

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 133    Accepted Submission(s): 67

Problem 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

Source

2015 Multi-University Training Contest 3

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5319

题目大意：刷墙，只能按两个斜45度刷，一个是红色，一个是蓝色，红遇蓝变绿，每个格子同种颜色只刷一次，每次刷必须是连续的一段，问答案目标状态最少刷几次

题目分析：按两个方向枚举，\这种样子刷，若当前点是红或绿且斜前一个点不是红且不是绿，则必然要刷一次，同理/这样刷时也判断一下，可是为什么这样算出来就是最小的呢，因为我只在必须要刷的时候才刷，所以显然这样就是最优的

#include <cstdio>
#include <cstring>
char s[55][55];

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

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