POJ2306 Lake Counting

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 37065 Accepted: 18433

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

题意：图中有多少个不同的连通块 比如左上角的算一个 左下角的算一个 右边的算一个 一共是3个

题解：dfs水题 hdu有题差不多的 符号换下就行了

#include <iostream>#include <stdio.h>#include <string.h>#include <queue>using namespace std;const int maxn = 105;char map[maxn][maxn];int move[8][2] = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};int sum,m,n;void dfs(int x, int y) {int tx;int ty;for (int i = 0; i<8; i++) {tx = x + move[i][0];ty = y + move[i][1];if (1 <= tx&&tx <= m && 1 <= ty&&ty <= n) {if (map[tx][ty] == 'W') {map[tx][ty] = '.';dfs(tx, ty);}}}}int main() {scanf("%d%d",&m,&n);sum=0;for (int i = 1; i<=m; i++) {for (int j = 1; j<=n; j++) {cin >> map[i][j];}}for (int i = 1; i<=m; i++) {for (int j = 1; j<=n; j++) {if (map[i][j] == 'W') {sum++;map[i][j] = '.';dfs(i,j);}}}cout << sum << endl;return 0;}