HDoj-1312-Red and Black -BFS

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10064    Accepted Submission(s): 6276

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<stdio.h>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int f[4][2]={{0,1},{0,-1},{1,0},{-1,0}};int t,m,n,X,Y;char str[21][21];void BFS(int x,int y){    int xx,yy,i;    if(x>=m||x<0||y<0||y>=n)      return;    for(i=0;i<4;i++)    {        xx=x+f[i][0];        yy=y+f[i][1];        if(xx<0||xx>=m||yy<0||yy>=n||str[xx][yy]=='#') continue;           t++;        str[xx][yy]='#'; //找出可以到达的最大区域，而不是路线，第二次竟然又不会了，呵呵哒，可以直接用‘#’覆盖走过的路。        BFS(xx,yy);    }}int main(){   int i,j;   while(~scanf("%d%d",&n,&m),m+n)   {      t=1;      getchar();      for(i=0;i<m;i++)      {         for(j=0;j<n;j++)         {            scanf("%c",&str[i][j]);            if(str[i][j]=='@')            {               X=i;               Y=j;            }          }            getchar();//keng      }       str[X][Y]='#';       BFS(X,Y);       printf("%d\n",t);   } return 0;}