Farm Irrigation
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 46 Accepted Submission(s) : 26
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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
看这这几张图有点懵，其实只要仔细看，也没有很难。因为他的连通就是一个十字形。分成I和一考虑
与上面有联系的是 ABEGHJK（及下面出头）
与下面有联系的是CDEHIJK
与左边有联系的是 ACFGHIK
与右边有联系的是BDFGIJK
#include <iostream> using namespace std;
char map[][];
int par[];
int find(int x)
{
while(x!=par[x])
x=par[x];
return x;
}
void unioni(int x,int y)
{
int xx=find(x);
int yy=find(y);
if(xx!=yy)
{
par[yy]=xx;
} }
int main()
{
int n,m;
while(cin>>n>>m)
{
if(n<||m<)
break;
for(int i=;i<n*m;i++)
par[i]=i;
for(int i=;i<n;i++)
for(int j=;j<m;j++)
{
cin>>map[i][j];
} for(int i=;i<n;i++)
for(int j=;j<m;j++)
{
if(i>=&&(map[i][j]=='A'||map[i][j]=='B'||map[i][j]=='E'||map[i][j]=='G'||map[i][j]=='H'||map[i][j]=='J'||map[i][j]=='K'))
{
if(map[i-][j]=='C'||map[i-][j]=='D'||map[i-][j]=='E'||map[i-][j]=='H'||map[i-][j]=='I'||map[i-][j]=='J'||map[i-][j]=='K')
{
unioni((i-)*m+j,i*m+j);
}
}
if(j>=&&(map[i][j]=='A'||map[i][j]=='C'||map[i][j]=='F'||map[i][j]=='G'||map[i][j]=='H'||map[i][j]=='I'||map[i][j]=='K'))
if(map[i][j-]=='B'||map[i][j-]=='D'||map[i][j-]=='F'||map[i][j-]=='G'||map[i][j-]=='I'||map[i][j-]=='J'||map[i][j-]=='K')
{
unioni(i*m+j-,i*m+j);
}
}
int flag=;
for(int i=;i<n*m;i++)
{
if(par[i]==i)
flag++; }
cout<<flag<<endl; }
return ;
}