POJ_Prob.ID:2386 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.
Input
- Line 1: The number of ponds in Farmer John’s field.
Output
For each test case, print a single integer indicating the minimum number of palantirs needed.
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.
Solution of source code
#include<stdio.h>
#include<iostream>
#define MAX_N 1000
//#define MAX_M 1000;
using namespace std;
int N,M;
char field[MAX_N][MAX_N];
void dfs(int x, int y){
field[x][y]='.';
for (int dx=-1;dx<=1;dx++){
for(int dy=-1;dy<=1;dy++){
int nx=x+dx,ny=y+dy;
if(0<=nx&&nx<N&&0<=ny&&ny<M&&field[nx][ny]=='W') dfs(nx,ny);
}
}
return;
}
int main(){
int res=0;
scanf("%d %d",&N,&M);
for(int i=0;i<N;i++){
for(int j=0;j<M;j++){
cin>>field[i][j];
}
}
for (int i=0;i<N;i++){
for(int j=0;j<M;j++){
if(field[i][j]=='W'){
dfs(i,j);
res++;
}
}
}
printf("%d\n",res);
}