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Farm Irrigation

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.

2 2
DK
HF

3 3
ADC
FJK
IHE

-1 -1

2
3

题意：告诉几个田，如果有管道联通则可以灌溉，问最少需要几个灌溉器

分析，主要就是求联通块，一个联通块一个灌溉器，直接求需要几个联通块就可以了。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
char s[1009][1009];
int vis[1009][1009];
int m,n;

int dis[4][2]={0,1,0,-1,1,0,-1,0};//右左下上
int up[]={1,1,0,0,1,0,1,1,0,1,1};
int dw[]={0,0,1,1,1,0,0,1,1,1,1};
int le[]={1,0,1,0,0,1,1,1,1,0,1};
int ri[]={0,1,0,1,0,1,1,0,1,1,1};

void dfs(int x,int y)
{
    if(vis[x][y]) return;
    vis[x][y]=1;

    for(int i=0;i<4;i++)
    {
        int xx=x+dis[i][0];
        int yy=y+dis[i][1];

        if(xx<0 || xx>=m || yy<0 || yy>=n)continue;
        if(i==0 && ri[s[x][y]-'A'] && le[s[xx][yy]-'A']) dfs(xx,yy);
        if(i==1 && le[s[x][y]-'A'] && ri[s[xx][yy]-'A']) dfs(xx,yy);
        if(i==2 && dw[s[x][y]-'A'] && up[s[xx][yy]-'A']) dfs(xx,yy);
        if(i==3 && up[s[x][y]-'A'] && dw[s[xx][yy]-'A']) dfs(xx,yy);
    }
}

int main()
{

    while(~scanf("%d%d",&m,&n))
    {
        if(n<0 || m<0 )break;
        memset(vis,0,sizeof vis);

        for(int i=0;i<m;i++)
            scanf("%s",s[i]);

        int ans=0;
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                if(vis[i][j]==0)
                {
                    ans++;
                    dfs(i,j);
                }

                printf("%d\n",ans);

    }
    return 0;
}

HDu 1198 Farm Irrigation

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原文地址：http://blog.csdn.net/wust_zjx/article/details/47984339