【简单并查集】Farm Irrigation

Farm Irrigation

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 3

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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 1Benny 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 2Several 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

题解：使用并查集，遍历每个格子的方向，把可以连通的相邻的两个格子合并成一个集合。输出N*M-Count就是独立集合的个数、

注意，并查集的数组需要开大点，开小了结果Tle了好几次。。。

代码：2015/10/20

#include <iostream>
#include <stdio.h>
#define Max 5200
using namespace std;
int DicX[4]={0,-1,0,1};
int DicY[4]={-1,0,1,0};
int Sign[15][4]={
1,1,0,0,//A
0,1,1,0,//B
1,0,0,1,//C
0,0,1,1,//D
0,1,0,1,//E
1,0,1,0,//F
1,1,1,0,//G
1,1,0,1,//H
1,0,1,1,//I
0,1,1,1,//J
1,1,1,1//K
};
int ID[Max];
char Str[Max][Max];
void Cread(int N)
{
    for(int i=0;i<=N;i++)ID[i]=i;
}
int Find(int x)
{
    if(x!=ID[x])ID[x]=Find(ID[x]);
    return ID[x];
}
int main()
{
    int N,M,i,j,k,d;
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        if(N<0||M<0)break;
        for(i=0;i<N;i++){
            scanf("%s",Str[i]);
        }
        int Count=0;
        Cread(N*M);
        for(i=0;i<N;i++)//遍历每一个格子
        {
            for(j=0;j<M;j++)
            {
                for(k=0;k<4;k++)//遍历当前格子的四个方向
                {
                    int ii=i+DicX[k];
                    int jj=j+DicY[k];
                    
                    if(ii<0||jj<0||ii>=N||jj>=M)continue;
                    if(Sign[Str[i][j]-'A'][k]&&Sign[Str[ii][jj]-'A'][(k+2)%4])
                    {//判断当前格子的四个方向与所相邻格子的是否可连通
                            int A=Find(i*M+j);
                            int B=Find(ii*M+jj);
                            if(A!=B)//合并可连通的集合
                            {
                                ID[A]=B;
                                Count++;
                            }
                    }
                }
            }
        }
        printf("%d
",N*M-Count);//输出集合的个数
    }
    return 0;
}