HDOJ 1312 Red and Black (DFS)

来源:互联网 发布:淘宝怎么投诉卖家客服 编辑:程序博客网 时间:2024/04/29 07:42

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14306    Accepted Submission(s): 8867


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
4559613
 

#include#include#includeint map[22][22];int w, h, max = INT_MIN;int dx[] = {-1,1,0,0},dy[] = {0,0,-1,1};void dfs(int x,int y){    int i;    for(i  =0; i < 4; i++)    {        if(!map[x+dx[i]][y+dy[i]]&&x+dx[i]<=h&&y+dy[i]<=w&&x+dx[i]>0&&y+dy[i]>0)        {            max++;            map[x+dx[i]][y+dy[i]]=1;            dfs(x+dx[i],y+dy[i]);        }    }}int main(){    int i,j,x,y;    char c;    while(scanf("%d %d",&w,&h)&&(w||h))    {        max=0;        memset(map,0,sizeof(map));        for(i=1;i<=h;i++)        {            getchar();            for(j=1;j<=w;j++)            {                scanf("%c",&c);                if(c=='.')                    map[i][j]=0;                else if(c=='#')                    map[i][j]=1;                else if(c=='@')                {                    map[i][j]=1;                    x=i;y=j;max=1;                }            }        }        dfs(x,y);        printf("%d\n",max);    }    return 0;}


0 0