题目链接
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
题目大意:
有一个大小为 N×M 的园子,雨后积起了水。八连通的积水被认为是连接在一起的。请求出 园子里总共有多少水洼?(八连通指的是下图中相对 W 的+的部分) +++ +W+ +++
题解:
用f[][] 记录被访问过的位置, 用dfs搜索
代码:
#include <iostream> #define MAX_N 110 #define MAX_M 110 using namespace std; char a[MAX_N][MAX_M]; bool f[MAX_N][MAX_M]; int N, M; int cnt = 0; int fx[] = {0, 1, 0, -1, 1, 1, -1, -1}; int fy[] = {1, 0, -1, 0, 1, -1, 1, -1}; void dfs(int x, int y) { f[x][y] = true; for (int i = 0; i < 8; i++) { int tx = x + fx[i]; int ty = y + fy[i]; if (!f[tx][ty] && a[tx][ty] == 'W') dfs(tx, ty); } } int main() { cin >> N >> M; for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) cin >> a[i][j]; for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { if (!f[i][j] && a[i][j] == 'W') { cnt++; dfs(i, j); } } } cout << cnt << endl; return 0; }