Lake Counting
Lake Counting
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
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
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Source Code:
#include <iostream> using namespace std; int go[][2]={ {-1,-1}, {-1,0}, {-1,1}, {0,-1}, {0,1}, {1,-1}, {1,0}, {1,1} }; const int arrSize=110; char maze[arrSize][arrSize]; void DFS(int x,int y,int n,int m){ maze[x][y]='.'; for(int i=0;i<8;++i){ int newX=x+go[i][0]; int newY=y+go[i][1]; if(newX<0||newX>=n) continue; if(newY<0||newY>=m) continue; if(maze[newX][newY]=='.') continue; DFS(newX,newY,n,m); } return; } int main() { int N,M; int pondsCount; while(cin>>N>>M){ for(int i=0;i<N;++i) for(int j=0;j<M;++j) cin>>maze[i][j]; pondsCount=0; for(int i=0;i<N;++i){ for(int j=0;j<M;++j){ if(maze[i][j]=='W'){ DFS(i,j,N,M); ++pondsCount; } } } cout<<pondsCount<<endl; } return 0; }