HDU1198 Farm Irrigation（并查集+模拟）

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12884    Accepted Submission(s): 5534

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

 

题目大意与分析

给出水管的种类，求出联通集的个数

判断每个格子能否和左边与上边的水管联通，如果能连通就加入并查集即可

注意二维数组与一维数组下标的转化问题，如果n行m列，那么一维数组下标是i*m+j

#include<bits/stdc++.h>
using namespace std;

char mp[55][55];
int n,m,i,j,s[5005],sum;

int pd_upanddown(int a,int b)
{
    int x1=a/m;
    int y1=a%m;
    int x2=b/m;
    int y2=b%m;
    if(mp[x1][y1]=='A'||mp[x1][y1]=='B'||mp[x1][y1]=='F'||mp[x1][y1]=='G')
    {
        return 0;
    }
    else if(mp[x2][y2]=='C'||mp[x2][y2]=='D'||mp[x2][y2]=='F'||mp[x2][y2]=='I')
    {
        return 0;
    }
    else
    {
        return 1;
    }
}

int pd_leftandright(int a,int b)
{
    int x1=a/m;
    int y1=a%m;
    int x2=b/m;
    int y2=b%m;
    if(mp[x1][y1]=='A'||mp[x1][y1]=='C'||mp[x1][y1]=='E'||mp[x1][y1]=='H')
    {
        return 0;
    }
    else if(mp[x2][y2]=='B'||mp[x2][y2]=='D'||mp[x2][y2]=='E'||mp[x2][y2]=='J')
    {
        return 0;
    }
    else
    {
        return 1;
    }
}

int findf(int x)
{
    return x==s[x]?x:s[x]=findf(s[x]);
}

void hebing(int a,int b)
{
    int fa=findf(a);
    int fb=findf(b);
    if(fa!=fb)
    {
        s[fb]=fa;
    }
}


int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        sum=0;
        if(m<0||n<0)
        {
            return 0;
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                s[i*m+j]=i*m+j;
            }
        }
        for(i=0;i<n;i++)
        {
            scanf("%s",&mp[i]);
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(i==0&&j==0)
                {
                    continue;
                }
                else if(i==0)
                {
                    if(pd_leftandright(i*m+j-1,i*m+j))
                    {
                        hebing(i*m+j-1,i*m+j);
                    }
                }
                else if(j==0)
                {
                    if(pd_upanddown(i*m+j-m,i*m+j))
                    {
                        hebing(i*m+j-m,i*m+j);
                    }
                }
                else
                {
                    if(pd_leftandright(i*m+j-1,i*m+j))
                    {
                        hebing(i*m+j-1,i*m+j);
                    }
                    if(pd_upanddown(i*m+j-m,i*m+j))
                    {
                        hebing(i*m+j-m,i*m+j);
                    }
                }
            }
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(findf(i*m+j)==(i*m+j))
                {
                    sum++;
                    //cout<<i<<" "<<j<<endl;
                }
            }
        }
        printf("%d
",sum);
    }
}