hdu 1312 Red and Black

Red and Black

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

解析：本题的意思是在不能翻墙（不能跳过‘#’）的情况下，最多能走遍多少个点

思路：还是一道经典的用dfs解决的题目，注意边界条件即可

#include<iostream>using namespace std;int cnt,a[4]={-1,0,1,0},b[4]={0,1,0,-1},n,m,vis[22][22];char s[22][22];void dfs(int x,int y){    for(int i=0;i<4;i++)    {        int p=x+a[i];        int q=y+b[i];        if(p>=0&&p<m&&q>=0&&q<n&&!vis[p][q]&&s[p][q]=='.')       //判断递归条件，包括在数组边界之内，该点未被标记         {            vis[p][q]=1;            //标记该点             cnt++;               //计数变量加一             dfs(p,q);               //递归搜索         }     }}int main(){    while(cin>>n>>m)    {        if(n==0&&m==0)          break;        cnt=0;        int r,l;        for(int i=0;i<m;i++)          for(int j=0;j<n;j++)          {               cin>>s[i][j];               vis[i][j]=0;               if(s[i][j]=='@')               {                    r=i;                    l=j;               }          }        cnt=1;          //赋初值         vis[r][l]=1;         //标记起点位置         dfs(r,l);          //递归搜索         cout<<cnt<<endl;    }}