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.
//大意就是许多方格农田呈矩形排布，每块农田上有各自的水管布局，水管能对接起来的农田间水可以互相流通
//求至少需要多少水源，能让水流遍所有农田
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 IHEthen 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
Source
本质上是很普通的DFS，但是加上那么多类型的水管就有些闹心了
打个表列举每种水管布局，剩下的就是裸DFS
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 // 7 #define RLQ main 8 // 9 using namespace std; 10 char map[100][100]; 11 int mx[5]={0,-1,0,1,0},//空 上右下左 12 my[5]={0,0,1,0,-1}; 13 int m,n; 14 int ans; 15 int pd(int ch,int i){ 16 switch(ch){ 17 case 'A':return i==1||i==4; 18 case 'B':return i==1||i==2; 19 case 'C':return i==3||i==4; 20 case 'D':return i==2||i==3; 21 case 'E':return i==1||i==3; 22 case 'F':return i==2||i==4; 23 case 'G':return i==1||i==4||i==2; 24 case 'H':return i==1||i==3||i==4; 25 case 'I':return i==2||i==3||i==4; 26 case 'J':return i==1||i==2||i==3; 27 case 'K':return 1; 28 } 29 } 30 void dfs(int x,int y){ 31 char ch=map[x][y]; 32 map[x][y]='M';//标记为已灌溉 33 int i; 34 for(i=1;i<=4;i++)//上 右 下 左 35 if(pd(ch,i)){//判断能否从这个方向出去 36 int x1=x+mx[i]; 37 int y1=y+my[i]; 38 if(x1>0 && x1<=m && y1>=0 && y1<n && map[x1][y1]!='M'){ 39 if(pd(map[x1][y1],(i+1)%4+1))//判断和下一个区块的联通 40 dfs(x1,y1); 41 } 42 } 43 return; 44 } 45 int RLQ(){ 46 int i,j; 47 while(scanf("%d%d",&m,&n) && m!=-1 && n!=-1){ 48 ans=0; 49 for(i=1;i<=m;i++)scanf("%s",map[i]); 50 for(i=1;i<=m;i++) 51 for(j=0;j<n;j++){ 52 if(map[i][j]!='M') ans++,dfs(i,j); 53 //TEST// for(int k=1;k<=m;k++)printf("%s ",map[k]); 54 } 55 printf("%d ",ans); 56 } 57 return 0; 58 }