杭电OJ——1198 Farm Irrigation （搜索）（2）

心已赠人 2021-09-26 11:34 201阅读 0赞

## 题目链接：[http://acm.hdu.edu.cn/showproblem.php?pid=1198][http_acm.hdu.edu.cn_showproblem.php_pid_1198] ##

## 题目： ##

# Farm Irrigation #

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.

![1121-1.gif][]

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

![1121-2.gif][]

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

## 题目描述： ##

时隔一段时间再写这道题又有了新的收获，看自己以前写的代码真的是有感触，

有图中几种形状的水管，给出一个n\*m的矩阵，问最少需要多少水源点可以使整个图有水流通。

一个搜索独立子图的问题，有意思的是水管形状各种各样，只有接口关联才可以流通，以前我写的是把各个水管的情况用if一一列举，然后搜索，这次学聪明了，用的二维数组存储的水管形状。要方便很多。

## 解题代码： ##

#include<stdio.h> #include<string.h> int a,b; char d[110][110]; int bo[110][110]; int qw[15][4] = { {1,1,0,0},{0,1,1,0},{1,0,0,1}, {0,0,1,1},{0,1,0,1},{1,0,1,0}, {1,1,1,0},{1,1,0,1},{1,0,1,1}, {0,1,1,1},{1,1,1,1} }; int next[4][2] = {0,-1, -1,0, 0,1, 1,0}; void dfs(int x,int y) { int tx,ty,i; if(bo[x][y]) return ; bo[x][y] = 1; for(i=0;i<4;i++){ tx = x + next[i][0]; ty = y + next[i][1]; if(tx >= 0 && tx < a && ty >= 0 && ty < b){ if(qw[d[x][y] - 'A'][i] && qw[d[tx][ty] - 'A'][(i + 2)%4]) //如果接口关联则搜索 dfs(tx,ty); } } return ; } int main() { int i,j; while(scanf("%d%d",&a,&b), a >= 0 && b >= 0){ memset(d,'\0',sizeof(d)); memset(bo,0,sizeof(bo)); for(i = 0 ;i < a;i ++) scanf("%s",d[i]); int t; for(t = i = 0 ;i < a;i ++){ for(j = 0;j < b;j ++){ if(!bo[i][j]){ t ++; dfs(i,j); } } } printf("%d\n",t); } return 0; }

[http_acm.hdu.edu.cn_showproblem.php_pid_1198]: http://acm.hdu.edu.cn/showproblem.php?pid=1198
[1121-1.gif]: /images/20210923/c3a89b68739743428719680a7e0566e6.png
[1121-2.gif]: /images/20210923/ec99ab2094e84e7985cc9b245b77c8ab.png