Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16042 | Accepted: 8118 |
Description
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
DFS基础题
将已经搜过的由'W'变成'.'，统计在这张图上一共进行过多少次DFS即为答案
因为忘了删freopen，还WA一次...
1 #include<iostream> 2 #include<cstdio> 3 4 using namespace std; 5 6 int n,m; 7 char garden[102][102]; 8 9 void dfs(int x,int y) 10 { 11 garden[x][y]='.'; 12 if(garden[x-1][y-1]=='W') 13 dfs(x-1,y-1); 14 if(garden[x-1][y]=='W') 15 dfs(x-1,y); 16 if(garden[x-1][y+1]=='W') 17 dfs(x-1,y+1); 18 if(garden[x][y-1]=='W') 19 dfs(x,y-1); 20 if(garden[x][y+1]=='W') 21 dfs(x,y+1); 22 if(garden[x+1][y-1]=='W') 23 dfs(x+1,y-1); 24 if(garden[x+1][y]=='W') 25 dfs(x+1,y); 26 if(garden[x+1][y+1]=='W') 27 dfs(x+1,y+1); 28 } 29 30 int main() 31 { 32 while(scanf("%d %d",&n,&m)==2) 33 { 34 int ans=0; 35 36 getchar(); 37 38 for(int i=1;i<=n;i++) 39 gets(&garden[i][1]); 40 for(int j=0;j<=m+1;j++) 41 garden[0][j]=garden[n+1][j]='.'; 42 for(int i=1;i<=n+1;i++) 43 garden[i][0]=garden[i][m+1]='.'; 44 45 for(int i=1;i<=n;i++) 46 for(int j=1;j<=m;j++) 47 if(garden[i][j]=='W') 48 { 49 ans++; 50 dfs(i,j); 51 } 52 53 printf("%d ",ans); 54 } 55 return 0; 56 }