hdu 5319 Painter（DFS）

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本文介绍了一种通过深度优先搜索(DFS)算法解决绘画板颜色转换问题的方法。问题旨在求解将绘画板上的格子从初始状态转换为指定状态所需的最小次数。绘画板由红色、蓝色和绿色格子组成，通过斜向绘制红蓝线条来改变颜色。

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Painter

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

Author

ZSTU

Source

2015 Multi-University Training Contest 3

ps：很水的DFS，主要就是英文太渣，不理解题意0.0

代码：

#include<stdio.h>
#include<string.h>

#define maxn 50+5
char s[maxn][maxn];
int vis[maxn][maxn];
int to1[][2]= {1,1,-1,-1},to2[][2]={-1,1,1,-1};
int n,m,ans;

void dfs1(int x,int y)//“\”的专属DFS
{
    for(int i=0; i<2; i++)
    {
        int xx=x+to1[i][0],yy=y+to1[i][1];
        if(xx>=0&&xx<n&&yy>=0&&yy<m)
        {
            if(s[xx][yy]=='R'&&!vis[xx][yy])
                vis[xx][yy]=1,dfs1(xx,yy);
            else if(s[xx][yy]=='G')
                s[xx][yy]='B',dfs1(xx,yy);
        }
    }
}

void dfs2(int x,int y)//“/”的专属DFS
{
    for(int i=0;i<2;i++)
    {
        int xx=x+to2[i][0],yy=y+to2[i][1];
        if(xx>=0&&xx<n&&yy>=0&&yy<m)
        {
            if(s[xx][yy]=='B'&&!vis[xx][yy])
                vis[xx][yy]=1,dfs2(xx,yy);
            else if(s[xx][yy]=='G')
                s[xx][yy]='R',dfs2(xx,yy);
        }
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for(int i=0; i<n; i++)
        {
            scanf("%s",s[i]);
            m=strlen(s[i]);
        }
        ans=0;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(s[i][j]=='R'&&!vis[i][j])
                {
                    ans+=1;
                    vis[i][j]=1;
                    dfs1(i,j);
                }
                else if(s[i][j]=='B'&&!vis[i][j])
                {
                    ans+=1;
                    vis[i][j]=1;
                    dfs2(i,j);
                }
                else if(s[i][j]=='G')
                {
                    ans+=2;
                    dfs1(i,j);
                    dfs2(i,j);
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}