POJ 2386

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 29838 Accepted: 14927

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 12W........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

<span style="font-family:Courier New;font-size:18px;"><strong>#include <iostream>#include <cstdio>using namespace std;#define MAX_N 110#define MAX_M 110int M, N;char field[MAX_M][MAX_N];void Input(){    cin >> M >> N;    for(int i=0; i<M; i++)        cin >> field[i];}void DFS(int x, int y){    field[x][y] = '.';    for(int move_x = -1; move_x <= 1; move_x++) {        for(int move_y = -1; move_y <=1 ; move_y++) {            if(!move_x && !move_y) continue;            int nx = x + move_x, ny = y + move_y;            if(0 <= nx && nx < M && 0 <= ny && ny < N && field[nx][ny] == 'W')  DFS(nx, ny);        }    }    return;}void Solve(){    int res = 0;    for (int i=0; i<M; i++) {        for(int j = 0; j<N; j++) {            if (field[i][j] == 'W') {                DFS(i, j);                res++;            }        }    }    printf("%d\n", res);}int main(){    Input();    Solve();    return 0;}/*10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.*/</strong></span>