Poj1979 Red and Black （DFS）

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 47466		Accepted: 25523

Description

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 adjacent tiles. But he can't move on red tiles, he can move only on black tiles. 

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20. 

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows. 

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 
The end of the input is indicated by a line consisting of two zeros. 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Japan 2004 Domestic

 

神坑的是：xy方向需要换一下

 

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
char a[35][35];
int n,m;
int res=0;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
void dfs(int x,int y)
{
    res++;
    a[x][y]='#';
    for(int i=0;i<4;i++){
        int nx=x+dx[i],ny=y+dy[i];
        if(nx>=0&&nx<n&&ny>=0&&ny<m&&a[nx][ny]=='.'){
            dfs(nx,ny);
        }
    }
    return ;
}
void solve()
{
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(a[i][j]=='@'){
                dfs(i,j);
            }
        }
    }
}
int main()
{
    while(cin>>m>>n&&n!=0&&m!=0){
        
        res=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>a[i][j];
            }
        } 
        solve();
        cout<<res<<endl;
    } 
    return 0;
}

 脱离参考书自己再根据自己的理解过一遍：

#include <iostream>
#include <algorithm> 
#include <cmath>
#include <string>
#include <cstring>
using namespace std;
int n,m;
char a[25][25];
int sx,sy,nx,ny;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int res;
void dfs(int x,int y)
{
    res++;
    a[x][y]='#';
    for(int i=0;i<4;i++){
        nx=x+dx[i],ny=y+dy[i];
        if(nx>=0&&nx<m&&ny>=0&&ny<n&&a[nx][ny]=='.'){
            dfs(nx,ny);
        }
    }
}
int main()
{
    while(cin>>n>>m&&(n&&m)){
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                cin>>a[i][j];
                if(a[i][j]=='@'){
                    sx=i,sy=j;
                }
            }
        }
        res=0;
        dfs(sx,sy);
        cout<<res<<endl;
    }    
    return 0;
}