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

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,

and one along the right side.

HINT

Source

Gold

USACO Gold 题，没有难度。。

#include <bits/stdc++.h>
using namespace std;
const int mx[8] = {1,1,1,0,0,-1,-1,-1};
const int my[8] = {1,0,-1,1,-1,1,0,-1};
int n,m,ans;
int a[110][110];
void dfs(int x,int y)
{
    a[x][y] = 0;
    for (int k = 0; k < 8; k ++)
    {
        int nx = mx[k] + x;
        int ny = my[k] + y;
        if (a[nx][ny]) dfs(nx,ny);
    }
}
inline int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}
int main()
{
    n = read();
    m = read();
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
        {
            char ch;
            cin >> ch;
            if(ch == 'W') a[i][j] = 1;
        }

    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++)
            if(a[i][j])
            {
                dfs(i, j);
                ans ++;
            }
    printf("%d",ans);
    return 0;
}