HDU 1198

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6933    Accepted Submission(s): 2966

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.

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 

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

//这题能够把图中的每个放格的连通性表示出来，以上下左右 能连通即为1反之为0  在dfs搜索

//本来想练下并查集 结果没想出来。。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
//上下左右
int n,m;  //矩阵大小
int vis[550][550];  //标记数组
int dir[4][2]={-1,0,1,0,0,-1,0,1};  //方向
int a[11][4]={
    1,0,1,0,
    1,0,0,1,
    0,1,1,0,
    0,1,0,1,
    1,1,0,0,
    0,0,1,1,
    1,0,1,1,
    1,1,1,0,
    0,1,1,1,
    1,1,0,1,
    1,1,1,1,
};    //11个字母的出口 上下左右 有出口为1 没有为0
struct Node
{
    int n;
    int a,b,c,d;  //上下左右
}ma[550][550];
void dfs(int xx,int yy)
{
    if(vis[xx][yy]) return;
    vis[xx][yy]=1;
    if((xx<0&&xx>=n)||(yy<0&&yy>=m))return;
    for(int i=0;i<4;i++)
    {
        int x1,y1;
        x1=xx,y1=yy;
        x1+=dir[i][0];
        y1+=dir[i][1];
        if(x1>=0&&x1<n&&y1>=0&&y1<m)  //边界
        {
            if(i==0&&(ma[xx][yy].a+ma[x1][y1].b==2))
            {
                dfs(x1,y1);
            }
            else if(i==1&&(ma[xx][yy].b+ma[x1][y1].a==2))
            {
                dfs(x1,y1);
            }
            else if(i==2&&(ma[xx][yy].c+ma[x1][y1].d==2))
            {
                dfs(x1,y1);
            }
            else if(i==3&&(ma[xx][yy].d+ma[x1][y1].c==2))
            {
                dfs(x1,y1);
            }
        }
    }

}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==-1&&m==-1) break;
        char c;
        for(int i=0;i<n;i++)
        {
            for(int k=0;k<m;k++)
            {
               cin>>c;
               ma[i][k].n=c-'A';
            }
        }
        int cnt=0;
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++)
            for(int k=0;k<m;k++)
                {
                    ma[i][k].a=a[ma[i][k].n][0],ma[i][k].b=a[ma[i][k].n][1];
                    ma[i][k].c=a[ma[i][k].n][2],ma[i][k].d=a[ma[i][k].n][3];
                }
        for(int i=0;i<n;i++)
        {
            for(int k=0;k<m;k++)
            {
                if(!vis[i][k])
                {
                    cnt++;
                    dfs(i,k);
                }
            }
        }
        printf("%d
",cnt);
    }
    return 0;
}