Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 46018 | Accepted: 22739 |
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.
#include<iostream>//将扫描到的W设置为. 是主要思想
using namespace std;
int n, m;
char x[100][100];
void dfs(int i, int j) {
x[i][j] = '.';
for (int a = -1; a <= 1; a++) {
for (int b = -1; b <= 1; b++) {
if (i + a < 0 || i + a >= n || j + b < 0 || j + b >= m)
continue;
if (x[i + a][j + b] == 'W') {
dfs(i + a, j + b);
}
}
}
}
int main() {
cin >> n >> m;
int cnt = 0;
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
cin >> x[i][j];
int i, j;
for (i = 0; i < n; i++) {
for (j = 0; j < m; j++) {
if (x[i][j] == 'W') {
dfs(i, j);
cnt++;
}
}
}
cout << cnt;
return 0;
}