Red and Black（DFS）

Red and Black

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

思路：借用DFS（点击打开链接）的知识点解答

#include<cstdio>  #include<iostream>#include<string.h>  #include<algorithm>  #define M 300  using namespace std;  int dx[4]={0,1,-1,0};//x的四个方向 上 下 左 右 int dy[4]={1,0,0,-1};//y的四个方向 上 下 左 右  int n, m, x, y, ans;   bool vis[M][M];  char map[M][M];  void F(int xx,int yy)  {      int i,j;      for(i=0;i<4;i++)//上下左右4个方向,所以4次      {   int xxx=xx+dx[i],  yyy=yy+dy[i];        if(map[xxx][yyy]=='.' && vis[xxx][yyy]==false )           //判断此位置是否为'.'   判断是否来过此位置         if( xxx>=0 && yyy>=0 && xxx<m && yyy<n ){ //判断目前的坐标是否超出范围            vis[xxx][yyy]=true;              ans++;                F(xxx,yyy);          }      }  }  int main()  {      while(scanf("%d %d",&n,&m)&&(n||m))      {             memset(vis,false,sizeof(vis));//把所有的点记为未走过的          for(int i=0;i<m;i++)          {              getchar();              for(int j=0;j<n;j++)              {                  scanf("%c",&map[i][j]);                  if(map[i][j]=='@')//标记@点的坐标                  {    x=i,y=j;    vis[i][j]=true;    }              }          }          ans=0;          F(x,y);          cout<<ans+1<<endl;//ans+1是因为@点也在到达范围内      }      return 0;  }