poj2386 Lake Counting(DFS解法)

/**********************
author: yomi
date: 18.8.14
ps:入门题果然很简单， 我觉得比选数，选物品什么的简单多了呀。
只要控制好递归边界和递归式就没问题。
**********************/
#include <iostream>
#include <queue>
#include<cstdio>
using namespace std;
int n, m, ans = 0;
const int maxn = 500;
bool vis[maxn][maxn];
char mmap[maxn][maxn];

void dfs(int x, int y)
{
    if(mmap[x][y] == '.')
        return;
    if(x<0 || x>=n || y<0 || y>=m || vis[x][y])
        return;
    vis[x][y] = true;
    dfs(x-1, y-1);
    dfs(x-1, y);
    dfs(x-1, y+1);
    dfs(x, y-1);
    dfs(x, y+1);
    dfs(x+1, y-1);
    dfs(x+1, y);
    dfs(x+1, y+1);

}
int main()
{
    cin >> n >> m;

    for(int i=0; i<n; i++){
        scanf("%s", mmap[i]);
    }

    for(int i=0; i<n; i++){
        for(int j=0; j<m; j++){
            if(mmap[i][j] == 'W' && !vis[i][j]){
                dfs(i, j);
                ans++;
            }
        }
    }

    cout << ans << endl;

    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.
**/
poj2386 Lake Counting(DFS解法)

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