hdu 1198(并查集判断连通性)

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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
   
   
   
   

 

解题思路：并查集判断连通性，注意要先预处理方块与方块直接的连通问题。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 55;
int n,m,land[11][4] = {{1,0,0,1},{1,1,0,0},{0,0,1,1},{0,1,1,0},
					   {1,0,1,0},{0,1,0,1},{1,1,0,1},{1,0,1,1},
					   {0,1,1,1},{1,1,1,0},{1,1,1,1}};
bool link[11][11][4],vis[maxn*maxn];
int fa[maxn*maxn];
char map[maxn][maxn];

void init()
{
	for(int i = 0; i < 11; i++)
		for(int j = 0; j < 11; j++)
		{
			if(land[i][0] == 1 && land[j][2] == 1)
				link[i][j][0] = true;
			if(land[i][1] == 1 && land[j][3] == 1)
				link[i][j][1] = true;
			if(land[i][2] == 1 && land[j][0] == 1)
				link[i][j][2] = true;
			if(land[i][3] == 1 && land[j][1] == 1)
				link[i][j][3] = true;
		}
}

int find(int x)
{
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}

void Union(int x,int y)
{
	int fx = find(x);
	int fy = find(y);
	if(fx != fy)
		fa[fy] = fx;
}

int direction(int x1,int y1,int x2,int y2) //(x2,y2)在(x1,y1)的哪个方向
{
	if(x1 == x2 + 1 && y1 == y2) return 0;
	else if(x1 == x2 && y1 + 1 == y2) return 1;
	else if(x1 + 1 == x2 && y1 == y2) return 2;
	else if(x1 == x2 && y1 == y2 + 1) return 3;
}

int main()
{
	int d;
	init();
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n == -1 && m == -1) break;
		for(int i = 0; i < n * m; i++) fa[i] = i;
		for(int i = 0; i < n; i++)
			scanf("%s",map[i]);
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++)
			{
				if(i > 0)
				{
					d = direction(i,j,i-1,j);
					if(link[map[i][j]-'A'][map[i-1][j]-'A'][d] == true)
						Union(i * m + j,(i - 1) * m + j);
				}
				if(j > 0)
				{
					d = direction(i,j,i,j-1);
					if(link[map[i][j]-'A'][map[i][j-1]-'A'][d] == true)
						Union(i * m + j,i * m + j - 1);
				}
			}
		int ans = 0;
		memset(vis,false,sizeof(vis));
		for(int i = 0; i < n * m; i++)
			if(vis[find(i)] == false)
			{
				vis[find(i)] = true;
				ans++;
			}
		printf("%d\n",ans);
	}
	return 0;
}