题目描述

blablabla

样例

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算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

//flood fill 连通问题 
//dfs实现 
#include<cstdio>
#include<iostream>

using namespace std;

const int N = 1010;
int n,m;
int res;
char field[N][N];
int dx[8]={1,0,-1,0,1,-1,1,-1},dy[8]={0,1,0,-1,1,-1,-1,1};

void dfs(int x, int y){
    field[x][y] = '.';
    int nx,ny;
    //9个方向实质上 0 0 就是自己本身. 所以和遍历八个方向是一样的.
    /*for(int dx = -1; dx <= 1; dx++){
        for(int dy = -1; dy <= 1; dy++){
            int nx = x + dx, ny = x + dy;
            if(0 <= nx && nx < n && 0 <= ny && ny < m && field[nx][ny] == 'W') dfs(nx,ny);
        }
    }
    然而测试的时候出错了？？？有人能告诉我为啥错了吗*/
    for(int i = 0; i < 8; ++i){
        nx = x + dx[i]; ny = y + dy[i];
        if(0 <= nx && nx < n && 0 <= ny && ny < m && field[nx][ny] == 'W') dfs(nx,ny);
    }

    return ;
}

void solve(){
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            if(field[i][j] == 'W'){
                dfs(i,j);
                res++;
            }
        }
    }
    cout << res << endl;
}

int main(){
    cin >> n >> m;
    for(int i = 0; i < n; i++) scanf("%s",field[i]);
    //for(int i = 0; i < n; i++) printf("测试 %s\n",field[i]);
    solve();
    return 0;
}