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
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2412
简单的dfs搜索，主要就是构图，将水管状态变为数组状态，可以将水管类型字母转化成3*3的方阵。
0,1,0, 0,1,0, 0,0,0, 0,0,0, 0,1,0, 0,0,0, 0,1,0, 0,1,0, 0,0,0, 0,1,0, 0,1,0,
1,1,0, 0,1,1, 1,1,0, 0,1,1, 0,1,0, 1,1,1, 1,1,1, 1,1,0, 1,1,1, 0,1,1, 1,1,1,
0,0,0, 0,0,0, 0,1,0, 0,1,0, 0,1,0, 0,0,0, 0,0,0, 0,1,0, 0,1,0, 0,1,0, 0,1,0
A B C D E F G H I J K
构图成功后就是简单的dfs。类似题目有HDOJ1241，HDOJ1312，POJ1562，POJ2386，POJ1979
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m;
int farm[155][155];
int dir[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int x[11][9]=
{
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
};
void build(char c,int signi,int signj);
void dfs(int x,int y);
int main()
{
int i,j;
char c;
while(scanf("%d%d",&n,&m)&&n>=0&&m>=0)
{
getchar();
memset(farm,0,sizeof(farm));
for(i=1; i<n*3-1; i+=3)
{
for(j=1; j<m*3-1; j+=3)
{
scanf("%c",&c);
build(c,i,j);
//cout<<i<<" "<<j<<endl;
}
getchar();
}
/*
cout<<endl;
for(i=0; i<n*3; i++)
{
for(j=0; j<m*3; j++)
printf("%d",farm[i][j]);
printf("
");
}
cout<<endl;
*/
int sum=0;
for(i=0; i<n*3; i++)
for(j=0; j<m*3; j++)
if(farm[i][j]==1)
{
farm[i][j]=0;
sum++;
dfs(i,j);
}
cout<<sum<<endl;
}
return 0;
}
void build(char c,int signi,int signj)
{
int i,j;
int k,t=0;
k=c-65;
for(i=signi-1; i<=signi+1; i++)
{
for(j=signj-1; j<=signj+1; j++)
{
farm[i][j]=x[k][t++];
//printf("%d",farm[i][j]);
}
//printf("
");
}
}
void dfs(int x,int y)
{
int i,fx,fy;
for(i=0; i<4; i++)
{
fx=x+dir[i][0];
fy=y+dir[i][1];
if(fx>=0&&fx<n*3&&fy>=0&&fy<m*3&&farm[fx][fy]==1)
{
farm[fx][fy]=0;
dfs(fx,fy);
}
}
}