H - Farm Irrigation

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本文介绍了一个农田灌溉问题的解决方法，通过构建3*3小矩阵来模拟不同类型的水管道，并使用深度优先搜索算法来确定最少所需的水源数量。

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.

<b< dd="">

Output

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

<b< dd="">

Sample Input

2 2
DK
HF

3 3
ADC
FJK
IHE

-1 -1

Sample Output

2
3

题意：一个农夫需要灌溉他的农田，每块农田管道连接方法不一样，现在给你一块n*n的矩形农田，如果农田之间的管道能够接通，这块农田就能被浇灌，如果不与其他农田连通，就需要新加一个水源，问最少需要几个水源。

思路：先将读取的矩阵每一块转换成一个3*3的小矩阵，然后构图，最后搜索即可。

提交错了一次，语言没看准...............

#include<stdio.h>
#include<string.h>
int a[500][500];
int m,n;
void dfs(int x,int y)
{
	int next[5][3]={{0,1},{1,0},{0,-1},{-1,0}};//方向数组
	int i,j,tx,ty;
	if(x<0||x>m*3||y<0||y>n*3)//越界判断
		return ;
	if(a[x][y]==1)
	{
	
		a[x][y]=0;//能够连通的点全部删除
		for(i=0;i<=3;i++)
		{ 
			dfs(x+next[i][0],y+next[i][1]);
		}
	}
		
}

int main()
{
	int t[50][5]={{0,1,0},{1,1,0},{0,0,0},{0,1,0},{0,1,1},{0,0,0},{0,0,0},{1,1,0},{0,1,0},
				  {0,0,0},{0,1,1},{0,1,0},{0,1,0},{0,1,0},{0,1,0},{0,0,0},{1,1,1},{0,0,0},
				  {0,1,0},{1,1,1},{0,0,0},{0,1,0},{1,1,0},{0,1,0},{0,0,0},{1,1,1},{0,1,0},
				  {0,1,0},{0,1,1},{0,1,0},{0,1,0},{1,1,1},{0,1,0}};//先将每种字符代表的管道用矩阵表
	int i,j,k;
	char s[60][60];
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		if(m==(-1)&&n==-1)
			break;
			
		memset(a,0,sizeof(a));
		memset(s,0,sizeof(s));
		for(i=1;i<=m;i++)
			scanf("%s",s[i]);
		
		int x,y;	
		for(i=1;i<=m;i++)
		{
			for(j=0;j<n;j++)
			{
				x=(i-1)*3;
			//	y=(j+1)*x;
				for(k=(s[i][j]-'A')*3;k<(s[i][j]-'A'+1)*3;k++)//构图
				{
					
			//		printf("s===%c k====%d\n",s[i][j],k);
					a[x][j*3]=t[k][0];
					a[x][j*3+1]=t[k][1];
					a[x][j*3+2]=t[k][2];
					x++;
				}
			}
		}
		
		int sum=0;
		for(i=0;i<m*3;i++)
		{
			for(j=0;j<n*3;j++)
			{
				if(a[i][j]==1)
				{
					dfs(i,j);//搜索
					sum++;
				}		
			}
		}
		printf("%d\n",sum);
	}					 				 
	return 0;
}