hdu1198 不一样的搜索水题

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.

题意   A--K对应不同水管     要事整个田地都能灌溉到水

问 需要到少水源          一开始想的就是深搜   只是方向都不同罢了    和求连通区域个数没什么区别    属于最水的搜索题

ac code

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;
const int maxn=60;
char mat[maxn][maxn];
bool vis[maxn][maxn];
int row,col;
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}
};
int dx[]= {-1,0,1,0};
int dy[]= {0,1,0,-1};

bool judge(int x,int y)
{
return x>0&&x<=row&&y>0&&y<=col&&!vis[x][y];
}

void dfs(int from,int x,int y)
{
int num=mat[x][y]-'A',xx,yy;
if(from<4)
if(!dir[num][(from+2)%4]){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);
}
}
}
}

int main()
{
while(scanf("%d%d",&row,&col),row>0&&col>0)
{
getchar();
for(int i=1; i<=row; i++)
scanf("%s",mat[i]+1);
memset(vis,false,sizeof vis);
int ans=0;
for(int i=1; i<=row; i++)
for(int j=1; j<=col; j++)
if(!vis[i][j])
{
ans++;
vis[i][j]=true;
dfs(maxn,i,j);
}
printf("%d\n",ans);
}
return 0;
}