1113. 红与黑

题意：

从@出发，问能走多少次“.”。每次只能走点。

思路：

简单走迷宫。

AC(抽风大佬的dfs)

# include <bits/stdc++.h>

using namespace std;
int ans=0, sx, sy;
char g[30][30];
int dx[]={
    
    0,0,1,-1};
int dy[]={
    
    1,-1,0,0};
void dfs(int x, int y){
    
    
    ans++;
    for(int i = 0; i < 4; i ++ ){
    
    
        int a = x+dx[i];
        int b = y + dy[i];
        if(g[a][b]=='.') g[a][b] = 0,dfs(a,b);
    }
}
int main(){
    
    
    int n, m;
    while(cin>>m>>n,n||m){
    
    
        ans = 0;
        memset(g,0,sizeof(g));
        for(int i = 1; i<= n; i ++ ){
    
    
            cin>> g[i]+1;
            for(int j = 1; j <= m; j ++ ){
    
    
                if(g[i][j] == '@')sx = i, sy =j;
            }
        }
        dfs(sx,sy);
        cout<<ans<<endl;
    }
    return 0;
}

AC（我的bfs，不用标记，直接改就好了）

# include <bits/stdc++.h>
# define mp make_pair
using namespace std;
typedef pair<int,int> pa;
int dx[]={
    
    0,1,-1,0};
int dy[]={
    
    1,0,0,-1};
int vis[30][30];
char g[30][30];
int main(){
    
    
    int n, m, sx, sy;
    while(cin>>m>>n){
    
    
    if(n==0&&m==0)break;
    memset(vis,0,sizeof(vis));
    for(int i = 1; i <= n; i ++ ){
    
    
        for(int j = 1; j <= m; j ++) {
    
    
            cin>>g[i][j];
            if(g[i][j]=='@'){
    
    
                sx = i, sy = j;
            }
        }
    }
    queue<pa>q;
    q.push(mp(sx,sy));
    int ans  = 0;
    while(!q.empty()){
    
    
        int x = q.front().first;
        int y = q.front().second;q.pop();
        if(vis[x][y]) continue;
        vis[x][y] = 1;
        ans++;
        
        for(int i = 0; i < 4; i ++ ){
    
    
            int tx = x + dx[i];
            int ty = y + dy[i];
            if(tx>=1 && tx<=n && ty>=1 && ty<= m && !vis[tx][ty] && g[tx][ty]=='.')q.push(mp(tx,ty));
        }
    }
    cout<<ans<<endl;
    }
    return 0;
}