zoj Farm Irrigation dfs

Farm IrrigationTime Limit: 2 Seconds Memory Limit: 65536 KBBenny 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 differenttype of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.Figure 1Benny 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 likeFigure 2Several 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 thissquare 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.InputThere 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.OutputFor each test case, output in one line the least number of wellsprings needed.Sample Input

2 2DKHF3 3ADC
FJK
IHE-1 -1

Sample Output

2
3

每个位置不是数字或字母，而是一张图片，这样的话只要用结构体来存每个位置的状态就行，每个方向上判断是否进行dfs。

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn=55;struct node{int up,down,left,right;node(){up=down=left=right=0;}};char ch[maxn][maxn];node map[maxn][maxn];bool vis[maxn][maxn];int n,m,tot;node change(char c){node t;if(c=='A'){t.up=1;t.left=1;return t;}else if(c=='B'){t.up=t.right=1;return t;}else if(c=='C'){t.left=t.down=1;return t;}else if(c=='D'){t.right=t.down=1;return t;}else if(c=='E'){t.up=t.down=1; return t;}else if(c=='F'){t.left=t.right=1;return t;}else if(c=='G'){t.up=t.left=t.right=1;return t;}else if(c=='H'){t.up=t.down=t.left=1;return t;}else if(c=='I'){t.left=t.right=t.down=1;return t;}else if(c=='J'){t.down=t.up=t.right=1;return t;}else {t.down=t.left=t.right=t.up=1;return t;}}bool out(int x,int y){return (x<0||x>=n||y<0||y>=m);}void dfs(int x,int y){if(map[x][y].up&&!out(x-1,y)&&!vis[x-1][y]&&map[x-1][y].down){vis[x-1][y]=1; dfs(x-1,y);}if(map[x][y].left&&!out(x,y-1)&&!vis[x][y-1]&&map[x][y-1].right){vis[x][y-1]=1; dfs(x,y-1);}if(map[x][y].down&&!out(x+1,y)&&!vis[x+1][y]&&map[x+1][y].up){vis[x+1][y]=1; dfs(x+1,y);}if(map[x][y].right&&!out(x,y+1)&&!vis[x][y+1]&&map[x][y+1].left){vis[x][y+1]=1; dfs(x,y+1);}}int main(){while(cin>>n>>m,n+m>0){for(int i=0;i<n;++i)scanf("%s",ch[i]);memset(vis,0,sizeof(vis));tot=0;for(int i=0;i<n;++i)for(int j=0;j<m;++j)map[i][j]=change(ch[i][j]);for(int i=0;i<n;++i)for(int j=0;j<m;++j){if(!vis[i][j]){vis[i][j]=1;dfs(i,j);tot++;}}cout<<tot<<endl;}return 0;}

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