• 欢迎光临~

# POJ--2386 Lake Counting(DFS)

23:18 2022-12-25

http://poj.org/problem?id=2386

reference:《挑战程序设计竞赛(第2版)》2.1.4 p33

### 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

``````#include<cstdio>
int N, M;
#define MAX_N 10000
#define MAX_M 10000
char field[MAX_N][MAX_M + 1];

//现在的位置是x, y
void dfs(int x, int y) {
field[x][y] = '.'; // 将现在的位置替换为.

//遍历8个方向
for (int dx = -1; dx <= 1; dx++) {
for (int dy = -1; dy <= 1; dy++) {
//移动的结果是nx,ny
int nx = x + dx, ny = y + dy;
//判断nx,ny是否再园子里且是否有积水
if (0 <= nx && nx < N && 0 <= ny && ny < M && field[nx][ny] == 'W' ) dfs(nx, ny);
}
}
return ;
}

void solve() {
int res = 0;
for(int i = 0; i < N; i++) {
for(int j = 0; j < M; j++){
if(field[i][j] == 'W') {
//从有W的地方开始dfs
dfs(i, j);
res++;
}
}
}
printf("%dn", res);
}

int main() {
while (~scanf("%d %dn", &N, &M)) {
for(int i = 0; i < N; i++) {
for(int j = 0; j < M; j++){
scanf("%c", &field[i][j]);
}
getchar();
}
solve();
}
// while(~scanf("%d %d",&N,&M)){
//     for(int i=0;i<=N-1;i++){
//         scanf("%s",field[i]);
//     }
//     solve();
// }
}``````