L - Farm Irrigation

最新推荐文章于 2019-07-24 08:32:39 发布

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最新推荐文章于 2019-07-24 08:32:39 发布

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分类专栏：并查集最小生成树文章标签：最小生成树并查集

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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

#include <cstdio>
#include <cstring>
char map[60][60];
int d[3600];
int path[11][4]=           //上下左右为顺序，1表示有水管，0表示没有 
   { {1,0,1,0},{1,0,0,1},{0,1,1,0},{0,1,0,1},{1,1,0,0},{0,0,1,1},{1,0,1,1},{1,1,1,0},{0,1,1,1},{1,1,0,1},{1,1,1,1} };
int n,m,sum;
int find(int x)
{
    while(x!=d[x]) 
       x=d[x];
    return x;
}
int merge(int x,int y)                 //如果并入一个集合返回1，sum-- 
{
    int fx=find(x);
    int fy=find(y);
    if(fx==fy) return 0;
    d[fx]=fy;
    return 1;
}
void search(int x,int y)
{
    if(x>0 && path[map[x][y]-'A'][0] && path[map[x-1][y]-'A'][1])   //该位置的"上"和同列上一行位置的"下"都是1时相连 
        sum-=merge(x*m+y,(x-1)*m+y);
    if(y>0 && path[map[x][y]-'A'][2] && path[map[x][y-1]-'A'][3])   //该位置的"左"和同行前一列位置的"右"都是1时相连 
        sum-=merge(x*m+y,x*m+y-1);
}
int main()
{
    while(scanf("%d%d",&n,&m),n!=-1&&m!=-1)
    {
    
    	for(int i=0;i<3600;i++)
    	   d[i]=i;
 	   
        for(int i=0;i<n;i++)  
            scanf("%s",map[i]);
        
        sum=n*m;                    //sum初始化为n*m表示每个正方形为一个集合 
       for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            {
                search(i,j);       //重点与技巧 
            }

        printf("%d\n",sum);
    } 
}