Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7427 Accepted Submission(s): 3185


Problem Description Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.


Figure 1

Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map

ADC
FJK
IHE

then the water pipes are distributed like


Figure 2

Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn.

Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him?

Note: In the above example, at least 3 wellsprings are needed, as those red points in Figure 2 show.

Input There are several test cases! In each test case, the first line contains 2 integers M and N, then M lines follow. In each of these lines, there are N characters, in the range of 'A' to 'K', denoting the type of water pipe over the corresponding square. A negative M or N denotes the end of input, else you can assume 1 <= M, N <= 50.

Output For each test case, output in one line the least number of wellsprings needed.

Sample Input

2 2
DK
HF

3 3
ADC
FJK
IHE

-1 -1

Sample Output

2
3

 

題意：告訴幾個田，如果有管道聯通則可以灌溉，問最少需要幾個灌溉器

分析，主要就是求聯通塊，一個聯通塊一個灌溉器，直接求需要幾個聯通塊就可以了。

 

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
char s[1009][1009];
int vis[1009][1009];
int m,n;

int dis[4][2]={0,1,0,-1,1,0,-1,0};//右左下上
int up[]={1,1,0,0,1,0,1,1,0,1,1};
int dw[]={0,0,1,1,1,0,0,1,1,1,1};
int le[]={1,0,1,0,0,1,1,1,1,0,1};
int ri[]={0,1,0,1,0,1,1,0,1,1,1};

void dfs(int x,int y)
{
    if(vis[x][y]) return;
    vis[x][y]=1;

    for(int i=0;i<4;i++)
    {
        int xx=x+dis[i][0];
        int yy=y+dis[i][1];

        if(xx<0 || xx>=m || yy<0 || yy>=n)continue;
        if(i==0 && ri[s[x][y]-'A'] && le[s[xx][yy]-'A']) dfs(xx,yy);
        if(i==1 && le[s[x][y]-'A'] && ri[s[xx][yy]-'A']) dfs(xx,yy);
        if(i==2 && dw[s[x][y]-'A'] && up[s[xx][yy]-'A']) dfs(xx,yy);
        if(i==3 && up[s[x][y]-'A'] && dw[s[xx][yy]-'A']) dfs(xx,yy);
    }
}

int main()
{

    while(~scanf(%d%d,&m,&n))
    {
        if(n<0 || m<0 )break;
        memset(vis,0,sizeof vis);

        for(int i=0;i