Red and Black

Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 49   Accepted Submission(s) : 30

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Problem 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)

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

Asia 2004, Ehime (Japan), Japan Domestic

#include<iostream>
#include<algorithm>
using namespace std;
int fang[4][2]={1,0,0,1,-1,0,0,-1};
bool mapused[25][25];
char map[25][25];
int n,m;//列 行；
int number;
int dfs(int a,int b)
{
    if(a>=0&&a<=m&&b>=0&&b<=n)
    {
        
          //判断有没有越界
           if(map[a][b]=='#')//遇墙 返回
           {
                return 0;
           }
       else
           if(mapused[a][b]==1)//遇到走过的 返回
          {
            return 0;
           }
        else//没走过 且不是墙；
        {
            mapused[a][b]=1;
            number++;
            int k;
            for(k=0;k<4;k++)//搜索4个方向
            {
                dfs(a+fang[k][0],b+fang[k][1]);
            }
    
    
        }

    }//越界结束
    else return 0;



}
int main()
{
    
    
    
    while(scanf("%d %d",&n,&m)!=EOF)//列 行
    {
        if(m==0&&n==0) break;
        memset(mapused,0,sizeof(mapused));
        number=0;
        memset(map,'#',sizeof(map));//初始化
        int i,j;
        int starx,stary;
        for(i=0;i<m;i++)
        {
            
            for(j=0;j<n;j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='@')
                {
                    starx=i;//行
                    stary=j;//列
                    
                
                }
            
            }
        
        }//输入地图；
        dfs(starx,stary);
        printf("%d
",number);
        

    
    
    
    
    }





}

这题广搜 深搜都可以

要注意初始化 和越界判断

每走过一步就 用mapused[] 标记已走过的地方 避免重复走

这题 起点重@开始 算能走多少格