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POJ 1979 Red and Black

2024-07-05 05:14:25   239人阅读

Red and Black
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 21482 Accepted: 11488

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

Source

Japan 2004 Domestic
思路　：简单的深搜
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int r,c;
char a[200][200];
int vis[200][200];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int sum;

void dfs(int i,int j)
{
             
  //if(a[i][j]==‘#‘||(vis[i][j])||(i<0||j<0||(i>r-1||j>c-1)))
  //return;
  //else 
  //{
      a[i][j]=‘#‘;  
      sum++;                                           
  for(int k = 0;k < 4;k++)
  {
         int x = i+dir[k][0];
         int y = j+dir[k][1];
         if(a[x][y]==‘.‘&&!vis[x][y]&&x>=0&&y>=0&&x<=r-1&&y<=c-1)  
        {
           dfs(x,y);
           vis[x][y]=1;
         }
  } 
  //} 
  
  
}
int main()
{
   int i,j,x,y;
   while(scanf("%d%d",&c,&r),r||c) 
   { 
          sum=0;
          memset(vis,0,sizeof(vis));
          memset(a,0,sizeof(a));
                                   
      for(i=0;i<r;i++)
      {
           getchar();
        for(j=0;j<c;j++)
        {
             a[i][j] = getchar();
             if(a[i][j]==‘@‘)
             {
                x=i;
                y=j;                
             }          
        }
       } 
       //count=0;
       dfs(x,y); 
       cout<<sum<<endl;                               
   }         
}



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