ZOJ - 2412 Farm Irrigation

Farm Irrigation

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

Submit Status

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

Source

Zhejiang University Local Contest 2005

 

 

 

 

分析：

图的表示+dfs。模板题。

ac代码：

//zoj2412
#include <iostream>
#include<cstdio>
using namespace std;
struct node
{
    int is;
    int left1,right1,up1,down1;
}mat[55][55];
struct node Change(char c)
{
    struct node t;
    t.down1=t.left1=t.right1=t.up1=0;
    if(c=='A')
    {
        t.left1=t.up1=1;
        return t;
    }
    if(c=='B')
    {
        t.right1=t.up1=1;
        return t;
    }
    if(c=='C')
    {
        t.left1=t.down1=1;
        return t;
    }
    if(c=='D')
    {
        t.right1=t.down1=1;
        return t;
    }
    if(c=='E')
    {
        t.up1=t.down1=1;
        return t;
    }
    if(c=='F')
    {
        t.left1=t.right1=1;
        return t;
    }
    if(c=='G')
    {
        t.left1=t.up1=t.right1=1;
        return t;
    }
    if(c=='H')
    {
        t.left1=t.up1=t.down1=1;
        return t;
    }
    if(c=='I')
    {
        t.left1=t.right1=t.down1=1;
        return t;
    }
    if(c=='J')
    {
        t.right1=t.up1=t.down1=1;
        return t;
    }
    if(c=='K')
    {
        t.left1=t.right1=t.up1=t.down1=1;
        return t;
    }
    //return t;//
}
void dfs(int x,int y)
{
    if(mat[x][y].up1&&mat[x-1][y].down1&&!mat[x-1][y].is)//!mat[x-1][y].is的判断很重要，否则出现Segmentation Fault （段错误），即堆栈耗费太大
    {
        mat[x-1][y].is=1;
        dfs(x-1,y);
    }
    if(mat[x][y].down1&&mat[x+1][y].up1&&!mat[x+1][y].is)
    {
        mat[x+1][y].is=1;
        dfs(x+1,y);
    }
    if(mat[x][y].left1&&mat[x][y-1].right1&&!mat[x][y-1].is)
    {
        mat[x][y-1].is=1;
        dfs(x,y-1);
    }
    if(mat[x][y].right1&&mat[x][y+1].left1&&!mat[x][y+1].is)
    {
        mat[x][y+1].is=1;
        dfs(x,y+1);
    }
}
int main()
{
    int i,j,m,n;
    char c;
    int count;
    while(scanf("%d%d",&m,&n)&&m>=0&&n>=0)
    {
        getchar();//读掉换行
        count=0;
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=n;j++)
            {
                scanf("%c",&c);
                //printf("%c ",c);//
                mat[i][j]=Change(c);//把Change(c)的返回值赋给mat[i][j]
                mat[i][j].is=0;
                //cout<<mat[i][j].down1<<mat[i][j].left1<<" ";//通过这个找到了错误
            }
            getchar();
        }
        for(i=1;i<=m;i++)//
        {
            mat[i][1].left1=mat[i][n].right1=0;
        }
        for(i=1;i<=n;i++)//
        {
            mat[1][i].up1=mat[m][i].down1=0;
        }
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(!mat[i][j].is)
                {
                    mat[i][j].is=1;
                    dfs(i,j);
                    count++;
                }
            }
        }
        printf("%d\n",count);
    }
    return 0;
}