Poj2386 Lake counting (DFS)

Starting from any W, the adjacent part is replaced. After one DFS operation, all the WS connected to the initial W will be replaced with '.', so until w no longer exists in the figure, the total number of DFS operations is the answer. Eight statuses are transferred in eight directions. Each grid is called once as a DFS parameter, so the complexity is O (8 × n × m) = O (n × m ).

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char map[110][110];int dis[8][2]={{0,1},{1,0},{-1,0},{0,-1},{1,1},{-1,-1},{-1,1},{1,-1}};int n,m;void dfs(int x,int y){    map[x][y]='.';    for(int i=0;i<8;i++)    {        int xx=x+dis[i][0];        int yy=y+dis[i][1];        if(xx<0||xx>=n||yy<0||yy>=m) continue;        if(map[xx][yy]=='W')        {            dfs(xx,yy);        }    }    return;}int main(){    //freopen("d:\\test.txt","r",stdin);    while(cin>>n>>m)    {        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                cin>>map[i][j];            }        }        int ans=0;        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                if(map[i][j]=='.') continue;                dfs(i,j);                ans++;            }        }        cout<<ans<<endl;    }    return 0;}

Poj2386 Lake counting (DFS)