poj 1979 Red and Black

关注

原创最新推荐文章于 2026-01-08 06:30:00 发布 330 阅读 AI 写同款· GEO 优化 ›

【题解】POJ 1979 Red and Black(DFS) POJ 1979 Red and Black 原题 There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four ... 阅读详情

（1）类型：深搜；
（2）题意：从‘@’出发，搜索可以到达的‘.’的个数；
（3）时间复杂度：O（W*H）；
（4）代码：

#include<iostream>
#include<cstring>
char field[26][26];
int n,m,res;
using namespace std;
void dfs(int x,int y){
    if(x<=n-1&&x>=0&&y>=0&&y<=m-1&&field[x][y]=='.'){field[x][y]='#';res++;}
    else return ;
    dfs(x-1,y);dfs(x,y-1);dfs(x,y+1);dfs(x+1,y);
}
int main(){
    ios::sync_with_stdio(false);cin.tie(0);
    int x,y;
    while(cin>>m>>n,n||m){
        res=0;memset(field,0,sizeof(field));
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++){
                cin>>field[i][j];
                if(field[i][j]=='@'){
                    x=i;y=j;
                    field[i][j]='.';
                }
            }
        dfs(x,y);
        cout<<res<<endl;
    }
}