HDU 1198 Farm Irrigation（DFS）

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9661    Accepted Submission(s): 4225

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

 

Author

ZHENG, Lu

 

Source

Zhejiang University Local Contest 2005

 

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Statistic | Submit | Discuss | Note

先把已知條件變成圖再進行DFS

如A變成         E變成         I變成                       

101                 101            111                           

001                 101            000                           

111                 101            101                     

即0爲通路，1爲不通路。油田  變形                    

變圖的過程一言難盡。。。每3*3部分四個角都爲1，然後再設5個位置的0/1

邊界都要*3

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
char map[200][200];
int nn,mm;
int dir[4][2]={1,0,-1,0,0,-1,0,1};
void find(char a,int b,int c){
	int i,j;
	map[b][c]='1';map[b+2][c]='1';map[b][c+2]='1';map[b+2][c+2]='1';
	switch(a){
		case 'A':map[b][c+1]='0';map[b+1][c]='0';map[b+1][c+1]='0';map[b+1][c+2]='1';map[b+2][c+1]='1';break;
		case 'B':map[b][c+1]='0';map[b+1][c]='1';map[b+1][c+1]='0';map[b+1][c+2]='0';map[b+2][c+1]='1';break;
		case 'C':map[b][c+1]='1';map[b+1][c]='0';map[b+1][c+1]='0';map[b+1][c+2]='1';map[b+2][c+1]='0';break;
		case 'D':map[b][c+1]='1';map[b+1][c]='1';map[b+1][c+1]='0';map[b+1][c+2]='0';map[b+2][c+1]='0';break;
		case 'E':map[b][c+1]='0';map[b+1][c]='1';map[b+1][c+1]='0';map[b+1][c+2]='1';map[b+2][c+1]='0';break;
		case 'F':map[b][c+1]='1';map[b+1][c]='0';map[b+1][c+1]='0';map[b+1][c+2]='0';map[b+2][c+1]='1';break;
		case 'G':map[b][c+1]='0';map[b+1][c]='0';map[b+1][c+1]='0';map[b+1][c+2]='0';map[b+2][c+1]='1';break;
		case 'H':map[b][c+1]='0';map[b+1][c]='0';map[b+1][c+1]='0';map[b+1][c+2]='1';map[b+2][c+1]='0';break;
		case 'I':map[b][c+1]='1';map[b+1][c]='0';map[b+1][c+1]='0';map[b+1][c+2]='0';map[b+2][c+1]='0';break;
		case 'J':map[b][c+1]='0';map[b+1][c]='1';map[b+1][c+1]='0';map[b+1][c+2]='0';map[b+2][c+1]='0';break;
		case 'K':map[b][c+1]='0';map[b+1][c]='0';map[b+1][c+1]='0';map[b+1][c+2]='0';map[b+2][c+1]='0';break;
	}
}
void dfs(int a,int b){
	int aa,bb,i;
	for(i=0;i<4;i++){
		aa=a+dir[i][0];
		bb=b+dir[i][1];
		if(aa>=0&&aa<nn&&bb>=0&&bb<mm&&map[aa][bb]=='0'){
			map[aa][bb]='1';
			dfs(aa,bb);
		}
	}
}
int main(){
	int i,j,n,m,ii,jj,sum;
	char ch;
	while(scanf("%d%d",&n,&m)&&n!=-1&&m!=-1){
		memset(map,0,sizeof(map));
		getchar();
		ii=0;
		for(i=0;i<n;i++){
			jj=0;
			for(j=0;j<m;j++){
				scanf("%c",&ch);
				find(ch,ii,jj);
				jj=jj+3; 
			}
			ii=ii+3;
			getchar();
		}
		nn=n*3; mm=m*3;
		/*
		printf("可以輸出這個矩陣觀察一下");
		for(i=0;i<nn;i++){
			for(j=0;j<mm;j++){
				printf("%c",map[i][j]);
			}
			printf("\n");
		}  printf("********\n"); 
		*/
		sum=0;
		for(i=0;i<nn;i++){
			for(j=0;j<mm;j++){
				if(map[i][j]=='0'){
					map[i][j]=='1';
					dfs(i,j);
					sum++;
				}
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}