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 * 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. 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 Output3
题意：要你求有多少个小水洼 简单的DFS题
#include<iostream> using namespace std; char a[110][110]; int n,m,b[8][2]={{0,1},{0,-1},{-1,0},{1,0},{-1,1},{1,1},{-1,-1},{1,-1}}; void DFS(int i ,int j) { a[i][j]='.'; int x,y,k; for(k=0;k<8;k++) { x=i+b[k][0]; y=j+b[k][1]; if(x<n&&x>=0&&y<m&&y>=0&&a[x][y]=='W') { DFS(x,y); } } } int main() { int i,j,k; while(cin>>n>>m) { k=0; for(i=0;i<n;i++) cin>>a[i]; for(i=0;i<n;i++) for(j=0;j<m;j++) { if(a[i][j]=='W') { k++; DFS(i,j); } } cout<<k<<endl; } return 0; }
