POJ 2386 Lake Counting

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

USACO 2004 November

 

 

 

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;
}