Red and Black——个人c++解

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....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

 

Sample Output

4559613
#include<bits/stdc++.h>using namespace std;int a,b,sum;char q[20][20];void su(int x,int y){    q[x][y]='#';//将走过的改为'#'    if(x-1>=0&&q[x-1][y]=='.')//向上    {        sum++;        su(x-1,y);    }    if(x+1<b&&q[x+1][y]=='.')//向下    {        sum++;        su(x+1,y);    }    if(y-1>=0&&q[x][y-1]=='.')//向左    {        sum++;        su(x,y-1);    }    if(y+1<a&&q[x][y+1]=='.')//向右    {        sum++;        su(x,y+1);    }    //return ;利用递归后退，每走到无路可走就后退。}int main(){    int i,j,i1,j1;    while(cin>>a>>b,a!=0&&b!=0)    {        sum=1;    for(i=0;i<b;i++)    {        for(j=0;j<a;j++)        {            cin>>q[i][j];            if(q[i][j]=='@')//标记@的位置            {                i1=i;                j1=j;            }        }    }    su(i1,j1);    cout<<sum<<endl;    }    return 0;}