Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

用深搜做的，一次深搜一个还没搜过格子，把和它连通的格子都标记，搜索次数就是答案。

貌似我的代码老长了，网上代码都很短。。。继续努力！！

#include"stdio.h"
#include"string.h"
#include"vector"
#include"algorithm"
using namespace std;
#define N 55
#define LL __int64
char g[N][N];
int n,m,vis[N][N];
int judge(int x,int y)
{
    if(x>=0&&x<n&&y>=0&&y<m&&!vis[x][y])
        return 1;
    return 0;
}
int down(int i,int j) //判断当前管子下面有没有接口
{
    if(g[i][j]=='C'||g[i][j]=='D'||g[i][j]=='E'||g[i][j]=='H'||g[i][j]=='I'||g[i][j]=='J'||g[i][j]=='K')
       return 1;
    return 0;
}
int up(int i,int j)   //判断当前管子上面有没有接口
{
    if(g[i][j]=='A'||g[i][j]=='B'||g[i][j]=='E'||g[i][j]=='G'||g[i][j]=='H'||g[i][j]=='J'||g[i][j]=='K')
       return 1;
    return 0;
}
int left(int i,int j)
{
    if(g[i][j]=='A'||g[i][j]=='C'||g[i][j]=='F'||g[i][j]=='G'||g[i][j]=='H'||g[i][j]=='I'||g[i][j]=='K')
       return 1;
    return 0;
}
int right(int i,int j)
{
    if(g[i][j]=='B'||g[i][j]=='D'||g[i][j]=='F'||g[i][j]=='G'||g[i][j]=='I'||g[i][j]=='J'||g[i][j]=='K')
       return 1;
    return 0;
}
void dfs(int x,int y)
{
    vis[x][y]=1;
    if(g[x][y]=='A')
    {
        if(judge(x-1,y)&&down(x-1,y))
            dfs(x-1,y);
        if(judge(x,y-1)&&right(x,y-1))
            dfs(x,y-1);
    }
    else if(g[x][y]=='B')
    {
        if(judge(x-1,y)&&down(x-1,y))
            dfs(x-1,y);
        if(judge(x,y+1)&&left(x,y+1))
            dfs(x,y+1);
    }
    else if(g[x][y]=='C')
    {
        if(judge(x+1,y)&&up(x+1,y))
            dfs(x+1,y);
        if(judge(x,y-1)&&right(x,y-1))
            dfs(x,y-1);
    }
    else if(g[x][y]=='D')
    {
        if(judge(x+1,y)&&up(x+1,y))
            dfs(x+1,y);
        if(judge(x,y+1)&&left(x,y+1))
            dfs(x,y+1);
    }
    else if(g[x][y]=='E')
    {
        if(judge(x+1,y)&&up(x+1,y))
            dfs(x+1,y);
        if(judge(x-1,y)&&down(x-1,y))
            dfs(x-1,y);
    }
    else if(g[x][y]=='F')
    {
        if(judge(x,y+1)&&left(x,y+1))
            dfs(x,y+1);
        if(judge(x,y-1)&&right(x,y-1))
            dfs(x,y-1);
    }
    else if(g[x][y]=='G')
    {
        if(judge(x-1,y)&&down(x-1,y))
            dfs(x-1,y);
        if(judge(x,y+1)&&left(x,y+1))
            dfs(x,y+1);
        if(judge(x,y-1)&&right(x,y-1))
            dfs(x,y-1);
    }
    else if(g[x][y]=='H')
    {
        if(judge(x+1,y)&&up(x+1,y))
            dfs(x+1,y);
        if(judge(x-1,y)&&down(x-1,y))
            dfs(x-1,y);
        if(judge(x,y-1)&&right(x,y-1))
            dfs(x,y-1);
    }
    else if(g[x][y]=='I')
    {
        if(judge(x,y+1)&&left(x,y+1))
            dfs(x,y+1);
        if(judge(x,y-1)&&right(x,y-1))
            dfs(x,y-1);
        if(judge(x+1,y)&&up(x+1,y))
            dfs(x+1,y);
    }
    else if(g[x][y]=='J')
    {
        if(judge(x+1,y)&&up(x+1,y))
            dfs(x+1,y);
        if(judge(x-1,y)&&down(x-1,y))
            dfs(x-1,y);
        if(judge(x,y+1)&&left(x,y+1))
            dfs(x,y+1);

    }
    else if(g[x][y]=='K')
    {
        if(judge(x+1,y)&&up(x+1,y))
            dfs(x+1,y);
        if(judge(x-1,y)&&down(x-1,y))
            dfs(x-1,y);
        if(judge(x,y+1)&&left(x,y+1))
            dfs(x,y+1);
        if(judge(x,y-1)&&right(x,y-1))
            dfs(x,y-1);
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m),n!=-1||m!=-1)
    {
        for(i=0;i<n;i++)
            scanf("%s",g[i]);
        int ans=0;
        memset(vis,0,sizeof(vis));
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(!vis[i][j])
                {
                    dfs(i,j);
                    ans++;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

hdu 1198 Farm Irrigation （搜索或并查集）,布布扣,bubuko.com