A - Lake Counting
阿新 • • 發佈:2019-01-06
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.
Input
* 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
#include<iostream> using namespace std; char map[101][101]; int dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1}}; int n,m,num; int dfs(int x,int y) { int a,b,k; map[x][y]='.'; for(k=0;k<8;++k) { a=x+dir[k][0]; b=y+dir[k][1]; if(a<n&&a>=0&&b<m&&b>=0&&map[a][b]=='W') dfs(a,b); } return 1; } int main() { int i,j; while(cin>>n>>m) { num=0; for(i=0;i<n;++i) cin>>map[i]; for(i=0;i<n;++i) for(j=0;j<m;++j) if(map[i][j]=='W') num+=dfs(i,j); cout<<num<<endl; } }