Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 24578 | Accepted: 12407 |
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 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.
Source
1 #include<cstdio>
2 #include<cstring>
3 #include<iostream>
4 #include<stack>
5 #include<set>
6 #include<map>
7 #include<queue>
8 #include<algorithm>
9 using namespace std;
10 char Map[105][105];
11 int n,m;
12 void dfs(int x,int y){
13 Map[x][y]=‘.‘;
14 int i,j;
15 for(i=-1;i<2;i++){
16 int xx=x+i;
17 for(j=-1;j<2;j++){
18 int yy=y+j;
19 if(xx>=0&&yy>=0&&xx<n&&yy<m&&Map[xx][yy]==‘W‘){
20 dfs(xx,yy);
21 }
22 }
23 }
24 }
25 int main(){
26 //freopen("D:\\INPUT.txt","r",stdin);
27 scanf("%d %d",&n,&m);
28 int i,j,k;
29 for(i=0;i<n;i++){
30 scanf("%s",Map[i]);
31 }
32 int res=0;
33 for(i=0;i<n;i++){
34 for(j=0;j<m;j++){
35 if(Map[i][j]==‘W‘){
36 dfs(i,j);
37 res++;
38 }
39 }
40 }
41 printf("%d\n",res);
42 return 0;
43 }