POJ2386Lake Counting

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 34226 Accepted: 17020

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 p

onds 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

经典的深搜，当然广搜也可以。

#include<cstdio>int N,M,book[101][101];char s[101][101];void dfs(int x,int y){    int next[8][2]={{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};    int tx,ty;    s[x][y]='.';    for(int k=0;k<=7;k++)    {        tx=x+next[k][0];        ty=y+next[k][1];        if(tx<1||ty<1||tx>N||ty>M)            continue;        if(book[tx][ty]==0&&s[tx][ty]=='W')        {            book[tx][ty]=1;            dfs(tx,ty);        }    }    return;}int main(){    int sum=0;    scanf("%d%d",&N,&M);    getchar();    for(int i=1;i<=N;i++)    {        for(int j=1;j<=M;j++)        {            scanf("%c",&s[i][j]);        }        getchar();    }    for(int i=1;i<=N;i++)        for(int j=1;j<=M;j++)        {            if(s[i][j]=='W')            {                dfs(i,j);                sum++;            }        }    printf("%d",sum);    return 0;}