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.

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)

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).

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
13


这是一道bfs的搜索题，只要按照4个方向来搜索即可。
下面是代码：

#include<stdio.h>

#include<stdlib.h>

#include<iostream>
using namespace std;
char str[30][30];

int flag[30][30];

int count;

int w,h;

int dir[4][2]={1,0,0,-1,-1,0,0,1}; //遍历数组
void bfs(int x,int y)

{

     int i;

     int ex,ey;

     for(i=0;i<4;i++)

     {

          ex=x+dir[i][0];

          ey=y+dir[i][1];

          if(ex<0||ex>=h||ey<0||ey>=w)

            continue; //记住不是break，也不是return，这里如果越界了，就返回到下一次循环

          if(flag[ex][ey]==0)

          {

                flag[ex][ey]=1;

                count++;

                bfs(ex,ey);

          }

     }

}
int main()

{

     int i,l;

     int sx,sy;

     while(scanf("%d%d",&w,&h)!=-1)

     {

          if(w==0&&h==0)

              break;

          for(i=0;i<h;i++)

          {

               scanf("%s",str[i]);

                 for(l=0;str[i][l];l++)

                 {

                      if(str[i][l]=='@')

                      {

                           sx=i;

                           sy=l;

                           flag[i][l]=1;

                      }

                      if(str[i][l]=='.')

                          flag[i][l]=0;

                      else

                      flag[i][l]=1;

                 }

          }

          count=1;

          bfs(sx,sy);

          printf("%d\n",count);

     }

     return 0;

}







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