课程练习二-1016-Red and Black

来源：互联网发布：开山刀淘宝哪里有卖编辑：程序博客网时间：2024/05/17 01:19

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<br>....#.<br>.....#<br>......<br>......<br>......<br>......<br>......<br>#@...#<br>.#..#.<br>11 9<br>.#.........<br>.#.#######.<br>.#.#.....#.<br>.#.#.###.#.<br>.#.#..@#.#.<br>.#.#####.#.<br>.#.......#.<br>.#########.<br>...........<br>11 6<br>..#..#..#..<br>..#..#..#..<br>..#..#..###<br>..#..#..#@.<br>..#..#..#..<br>..#..#..#..<br>7 7<br>..#.#..<br>..#.#..<br>###.###<br>...@...<br>###.###<br>..#.#..<br>..#.#..<br>0 0<br>

Sample Output

45<br>59<br>6<br>13<br>
题意：有红、黑两种颜色，只能走黑色，上、下、左、右走。求走过的黑色的。
思路：dfs，注意走过的地方。
代码：
#include <iostream> #include<stdlib.h> #include<string.h>#include<fstream>using namespace std;    char a[25][25];  int flag[25][25];  int sum;    void DFS(int x,int y)  {      if(a[x-1][y]=='.' && flag[x-1][y]==0)      {          sum++;          flag[x-1][y]=1;          DFS(x-1,y);      }      if(a[x][y-1]=='.' && flag[x][y-1]==0)      {          sum++;          flag[x][y-1]=1;          DFS(x,y-1);      }      if(a[x][y+1]=='.' && flag[x][y+1]==0)      {          sum++;          flag[x][y+1]=1;          DFS(x,y+1);      }      if(a[x+1][y]=='.' && flag[x+1][y]==0)      {          sum++;          flag[x+1][y]=1;          DFS(x+1,y);      }  }    int main()  {      freopen ("C:\\Users\\liuzhen\\Desktop\\11.txt", "r", stdin);    int n,m,i,j;      while(cin>>m>>n)      {          if(m==0&&n==0)break;         sum=0;          memset(flag,0,sizeof(flag));          memset(a,'#',sizeof(a));          for(i=1;i<=n;i++)          {              for(j=1;j<=m;j++)              {                  cin>>a[i][j];              }          }          for(i=1;i<=n;i++)          {              for(j=1;j<=m;j++)              {                  if(a[i][j]=='@')                  {                      sum++;                      flag[i][j]=1;                      DFS(i,j);                      i=n;                      j=m;                  }              }          }          cout<<sum<<endl;      }      freopen ("con", "r", stdin);    system("pause");    return 0;  }

0 0