POJ 2386 Lake Counting DFS
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Lake Counting
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
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很水的一道dfs但是有一个诡异的错误
代码中 int dx=x-i; int dy=y-j; 是对的 但 int dx=x+i; int dy=y+j;就错了连样例都过不了无法理解www
ACcode：
#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 1011
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI acos(-1.0)
#define E exp(1)
using namespace std;
char aa[maxn][maxn];
int n,m;
void dfs(int x,int y){
aa[x][y]='.';
FOR(i,-1,1)
FOR(j,-1,1){
int dx=x-i;
int dy=y-j;
if(dx>=0&&dx<n&&dy>=0&&dy<m&&aa[dx][dy]=='W')
dfs(dx,dy);
}
return ;
}
int main(){
while(rd2(n,m)!=EOF){
FOR(i,0,n-1)
FOR(j,0,m-1)
cin>>aa[i][j];
int cnt=0;
FOR(i,0,n-1)
FOR(j,0,m-1)
if(aa[i][j]=='W'){
dfs(i,j);
cnt++;
}
printf("%d\n",cnt);
}
return 0;
}
/*
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
*/