Farm Irrigation

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

分析：

問你連通圖的個數，並查集，但是要自己構造連通條件

連通條件：

對每個方塊，看它的左邊和上面的方塊能否和它連通，能的話就合並

註意方塊的邊界，比如第0行，就只要看他左邊的方塊能不能和他合並，不用看上面，因為上面是空的

第0行第0列的方塊 跳過

第0行的 只看他左邊的

第0列的 只看他上面的

其他的 看左邊和上面的

11個方塊，一個方塊的四條邊有管子的就是1，比如A塊，1，1，0，0

#include<bits/stdc++.h>
using namespace std;
#define max_v 51
int t[11][4]={1,1,0,0,
0,1,1,0,
1,0,0,1,
0,0,1,1,
0,1,0,1,
1,0,1,0,
1,1,1,0,
1,1,0,1,
1,0,1,1,
0,1,1,1,
1,1,1,1,
};
int sum;
int pa[max_v*max_v];//數組大小需要註意，坑了很多次
int rk[max_v*max_v];
void make_set(int x)
{
    pa[x]=x;
    rk[x]=0;
}
int find_set(int x)
{
    if(x!=pa[x])
       pa[x]=find_set(pa[x]);
       return pa[x];
}
void union_set(int x,int y)
{
    x=find_set(x);
    y=find_set(y);
    if(x==y)
        return ;
    sum--;
    if(rk[x]>rk[y])
    {
        pa[y]=x;
    }else
    {
        pa[x]=y;
        if(rk[x]==rk[y])
            rk[y]++;
    }
}
int main()
{
    int n,m;
    int f[max_v][max_v];
    char c;
    while(~scanf("%d %d",&m,&n))
    {
        getchar();
        if(n==-1&&m==-1)
            break;
       // memset(f,0,sizeof(f));
        sum=n*m;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                scanf("%c",&c);
                f[i][j]=c-‘A‘;//轉換
                make_set(i*n+j);//初始化
                if(i==0&&j==0)
                    continue;
                else if(i==0)
                {
                    if(t[f[i][j-1]][2]&&t[f[i][j]][0])//第0行，判斷左
                        union_set(i*n+j,i*n+j-1);
                }
                else if(j==0)//第0列，判斷上
                {
                    if(t[f[i][j]][1]&&t[f[i-1][j]][3])
                        union_set(i*n+j,(i-1)*n+j);
                }else
                {
                    // 其他 判斷左和上
                    if(t[f[i][j-1]][2]&&t[f[i][j]][0])
                        union_set(i*n+j,i*n+j-1);
                    if(t[f[i][j]][1]&&t[f[i-1][j]][3])
                        union_set(i*n+j,(i-1)*n+j);
                }
            }
            getchar();
        }
        printf("%d\n",sum);
    }
    return 0;
}

HDU 1198 Farm Irrigation（並查集，自己構造連通條件）