HDU 1198 Farm Irrigation (并查集)

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5809    Accepted Submission(s): 2516

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

 

Recommend

Ignatius.L

 

这道题就是问给定的图中能构成几个连通块。

既然是求连通块，那么就可以用并查集来做，但是这里怎么去判断连通情况要考虑考虑

我假定每一个方块有四个方向，如果这个方向能延伸出去和其他方块连通的话，那么就给这个方向设定为1，否则为0

然后对每一个方块进行枚举，如果能连通就合并。

最后统计根节点个数就行了。

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN=55;
char ch;
int a[MAXN][MAXN],p[MAXN*MAXN];
int dir[4][2]={{-1,0},{0,1},{1,0},{0,-1}};//寻找方向:上右下左
int map[11][4]={{1,0,0,1},//从上右下左四个方向依次列出每一个方块的连通情况
                {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}
                };
int Find(int x)
{
    return p[x]==x?x:p[x]=Find(p[x]);
}

void Union(int a,int b)
{
    int x=Find(a);
    int y=Find(b);
    if(x!=y)
        p[x]=y;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n,m;
    while(scanf("%d %d%*c",&m,&n)&&m>0)
    {
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%c",&ch);
                a[i][j]=ch-'A';
            }
            getchar();
        }

        for(int i=0;i<n*m;i++)
            p[i]=i;

        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                for(int k=0;k<4;k++)
                {
                    int next_x,next_y;
                    next_x=i+dir[k][0];
                    next_y=j+dir[k][1];
                    if(0<=next_x&&next_x<m&&0<=next_y&&next_y<n)
                    {
                        if(k==0)
                        {
                            if(map[a[next_x][next_y]][2]&&map[a[i][j]][0])
                                Union(next_x*n+next_y,i*n+j);
                        }
                        if(k==1)
                        {
                            if(map[a[next_x][next_y]][3]&&map[a[i][j]][1])
                                Union(next_x*n+next_y,i*n+j);
                        }
                        if(k==2)
                        {
                            if(map[a[next_x][next_y]][0]&&map[a[i][j]][2])
                                Union(next_x*n+next_y,i*n+j);
                        }
                        if(k==3)
                        {
                            if(map[a[next_x][next_y]][1]&&map[a[i][j]][3])
                                Union(next_x*n+next_y,i*n+j);
                        }
                    }
                }
            }
        }
//        for(int i=0;i<n*m;i++)
//            printf("%d ",p[i]);
//        printf("
");

        int cnt=0;
        for(int i=0;i<n*m;i++)
            if(p[i]==i)
                cnt++;
        printf("%d
",cnt);
    }
}