DFS (Lake Counting)
Lake Counting
原題連結:http://poj.org/problem?id=2386
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
我所理解的題目意思和注意點:
沒啥可說的,比較容易,直接上程式碼(挑戰程式設計)。
#include<stdio.h> #define MAX_N 100 #define MAX_M 100 int N,M,res; char field[MAX_N][MAX_M]; int dfs(int x,int y) { field[x][y]='.'; for(int dx=-1;dx<=1;dx++){ for(int dy=-1;dy<=1;dy++){ int nx=x+dx; int ny=y+dy; if(0<=nx&&nx<N&&0<=ny&&ny<M&&field[nx][ny]=='W') dfs(nx,ny); } } } int main() { scanf("%d%d",&N,&M); for(int i=0;i<N;i++) scanf("%s",field[i]); for(int i=0;i<N;i++){ for(int j=0;j<M;j++){ if(field[i][j]=='W'){ dfs(i,j); res++; } } } printf("%d\n",res); return 0; }
emm.......祝我國慶快樂。