Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 28966 | Accepted: 14505 |
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
解析：DFS。从一个'W'开始，每次把'W'连通的部分消掉，经过多次这种操作之后，图中不再有'W'，操作的次数就是结果。
#include <cstdio> int n, m;
char s[105][105]; bool inField(int r, int c)
{
return r >= 0 && r < n && c >= 0 && c < m;
} void dfs(int x, int y)
{
s[x][y] = '.';
for(int i = -1; i <= 1; ++i){
for(int j = -1; j <= 1; ++j){
int tx = x+i, ty = y+j;
if(inField(tx, ty) && s[tx][ty] == 'W')
dfs(tx, ty);
}
}
} int main()
{
scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i)
scanf("%s", s[i]);
int res = 0;
for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j){
if(s[i][j] == 'W'){
++res;
dfs(i, j);
}
}
}
printf("%d\n", res);
return 0;
}