Farm Irrigation【DFS】

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本文探讨了一种基于图论的算法解决农场灌溉问题，通过分析不同类型的水管布局，确定最少的水源点数量以确保整个农场得到充分灌溉。利用深度优先搜索(DFS)遍历地图，识别独立的灌溉区域。

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

题意：有如上图11种土地块，块中的绿色线条为土地块中修好的水渠，现在一片土地由上述的各种土地块组成，需要浇水，问需要打多少口井。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100;
char mapp[maxn][maxn];
bool vis[maxn][maxn];
int M,N;
int dir[12][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}
};//上，右，下，左
int dx[]= {-1,0,1,0};
int dy[]= {0,1,0,-1};
int judge(int x,int y)
{
    if(x>0&&x<=M&&y>0&&y<=N&&!vis[x][y])
        return true;

    else
        return false;
}
void dfs(int p,int x,int y)
{
    int num=mapp[x][y]-'A';//求出这个块是第几个
    int xx,yy;
    if(p<4)
        if(!dir[num][(p+2)%4])//判断是否连通
        {
            vis[x][y]=false;
            return;
        }
    for(int i=0; i<4; i++)
    {
        if(dir[num][i])
        {
            xx=x+dx[i];
            yy=y+dy[i];
            if(judge(xx,yy))
            {
                vis[xx][yy]=true;
                dfs(i,xx,yy);//i表示方向。
            }
        }
    }



}


int main()
{
    while(cin>>M>>N)
    {
        if(M==-1&&N==-1)
            break;
        for(int i=1; i<=M; i++)
            for(int j=1; j<=N; j++)
                cin>>mapp[i][j];
        memset(vis,false,sizeof(vis));
        int res=0;
        for(int i=1; i<=M; i++)
            for(int j=1; j<=N; j++)
                if(!vis[i][j])//遍历每一个位置，找出连通区域的数量。
                {
                    vis[i][j]=true;
                    res++;
                    dfs(maxn,i,j);
                }
        cout<<res<<endl;
    }

    return 0;
}