POJ2386 Lake Counting 【DFS】
來源：程序員人生 發布時間：2014-11-14 08:23:31 閱讀次數：2347次
Lake Counting
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 20782 |
|
Accepted: 10473 |
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
USACO 2004 November
睡前水1水。
#include <stdio.h>
#include <string.h>
#define maxn 102
char G[maxn][maxn];
int n, m;
const int mov[][2] = {0, 1, 0, ⑴, 1, 0, ⑴,
0, 1, ⑴, ⑴, 1, 1, 1, ⑴, ⑴};
void DFS(int x, int y) {
G[x][y] = '.';
int i, j, nx, ny;
for(i = 0; i < 8; ++i) {
nx = x + mov[i][0];
ny = y + mov[i][1];
if(nx >= 0 && nx < n && ny >= 0 && ny < m && G[nx][ny] == 'W')
DFS(nx, ny);
}
}
int main() {
int i, j, ret;
while(scanf("%d%d", &n, &m) == 2) {
for(i = 0; i < n; ++i)
scanf("%s", G[i]);
ret = 0;
for(i = 0; i < n; ++i)
for(j = 0; j < m; ++j)
if(G[i][j] == 'W') {
DFS(i, j);
++ret;
}
printf("%d
", ret);
}
return 0;
}
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