hduoj1198(并查集)

Farm Irrigation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8591 Accepted Submission(s): 3691

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(表示) 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
题解：将农场中的方块从0到m*n进行编号，然后每一个方块与相邻的右边和下面的方块进行判定是否联通，如果联通则将两个方块放到一个集合中，对计数器减一，最后输出计数器也就是联通的集合数。

#include<iostream>#include<cstdio>using namespace std;int map[555][555];int n,m;int pre[555*555];int cnt;int linked[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}                    };void makeset(){    int i;    for(i=0;i<m*n;i++){        pre[i]=i;    }    cnt=m*n;    return;}int find(int x){    if(x!=pre[x]){        pre[x]=find(pre[x]);    }    return pre[x];} void uniun(int a,int b){    if(find(b)!=find(a)){        pre[find(b)]=find(a);        cnt--;    }}void count(int x1,int y1,int x2,int y2,bool r){    if(x2>m-1||y2>n-1)return;    bool mark=false;    //cout<<mark<<endl;    if(r){        if(linked[map[x1][y1]][1]&&linked[map[x2][y2]][3])mark=true;    }    else{        if(linked[map[x1][y1]][2]&&linked[map[x2][y2]][0])mark=true;    }    //cout<<mark<<endl;    if(mark){        uniun(x1*n+y1,x2*n+y2);        //uniun((x1-1)*m+y1,(x2-1)*m+y2);        //printf("%d\n",cnt);           //cnt--;    }    return; }int main(){    int i,j;    char c;    while(scanf("%d%d",&m,&n)){        if(n==-1&&m==-1)break;        makeset();        for(i=0;i<m;i++)            for(j=0;j<n;j++){                cin>>c;                map[i][j]=c-'A';                    }        for(i=0;i<m;i++)            for(j=0;j<n;j++){                //printf("map=%d ",map[i][j]);                count(i,j,i+1,j,false);                count(i,j,i,j+1,true);                  }        printf("%d\n",cnt);    }       return 0;}