POJ - 1979 Red and Black （dfs+染色）

POJ - 1979 Red and Black （dfs+染色）

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#### personally think that this question really comes down to count consecutive '.' Number, then the '#' blocked.
#### ideas: @ search to determine the starting position, to become @ '#'; and then look for the four directions to find them one by one into a '#'; define global variables to be counted on time, and finally print out the answer. '' emmm because there may be multiple examples so ans have to remember is initialized to 0;

#include <iostream>
using namespace std;

int w,h,ans=0;
char a[21][21];

void dfs(int i, int j){
    a[i][j]='#';ans++;
    //查找四个方向
    if (a[i+1][j]=='.'&&(i+1)<w)dfs(i+1,j);
    if (a[i-1][j]=='.'&&(i-1)>=0)dfs(i-1,j);
    if (a[i][j+1]=='.'&&(j+1)<h)dfs(i,j+1);
    if (a[i][j-1]=='.'&&(j-1)>=0)dfs(i,j-1);
    return ;
}
int main()
{
    while(cin >> h >> w)
    {
        if (w==0&&h==0) return 0;
        for (int i=0;i<w;i++){
            for(int j=0;j<h;j++){
                cin >> a[i][j];
            }
        }
        for (int i=0;i<w;i++){
            for(int j=0;j<h;j++){
                if (a[i][j]=='@') dfs(i,j);
            }
        }
        cout<< ans<<endl;
        ans=0;
    }
}

This algorithm was rather powerful, very powerful recursive, wrote out super happy ha ha ha, emmm after all, is just learning the day before yesterday, and today there is a beginning point forgotten ,, ashamed.
-2020.02.12