POJ 1979 Red And Black 红与黑DFS深搜 AC代码C++

Red and Black

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 53332 Accepted: 28261

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

AC代码：

#include<iostream>
using namespace std;
const int MAX_WH=20;
char m[MAX_WH+1][MAX_WH+1];
int w=1,h=1,sx,sy,ans=0,count=0,res[1000],mx[4]={-1,0,1,0},my[4]={0,-1,0,1};
int main()
{
 void solve();
 while(!(w==0&&h==0))
 {
  ans=0;
  cin>>w>>h;
  if(w==0&&h==0)  break;
  for(int i=0;i<h;i++)
   for(int j=0;j<w;j++)
    cin>>m[i][j];
  solve();
  count++;
 }
 for(int i=0;i<count;i++)
  cout<<res[i]<<endl;
 return 0;
}
void solve()
{
 void dfs(int x,int y);
 for(int i=0;i<h;i++)
  for(int j=0;j<w;j++)
   if(m[i][j]=='@')
    dfs(i,j);
 res[count]=ans;
}
void dfs(int x,int y)
{
 ans++;
 m[x][y]='#';
 for(int i=0;i<4;i++)
 {
  int nx=x+mx[i],ny=y+my[i];
  if(nx>=0&&nx<h&&ny>=0&&ny<w&&m[nx][ny]=='.')
   dfs(nx,ny);
 }
}

注意：

①输入w h对应 列 和 行
②题目要求连续输入 所以我用数组保存各个结果 count用来计数