POJ 1979 Red and Black

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 25014 Accepted: 13502

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

4559613

#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#define N 1009using namespace std;char s[N][N];int n,m;int sx,sy;int dis[4][2]={ {0,1},{0,-1},{1,0},{-1,0} };int sum;void dfs(int x,int y){    for(int i=0;i<4;i++)    {        int xx=x+dis[i][0];        int yy=y+dis[i][1];        if(xx>=0 && xx<n && yy>=0 && yy<m && s[xx][yy]=='.')        {           s[xx][yy]='#';           sum++;           dfs(xx,yy);        }    }}int main(){    while(scanf("%d %d",&m,&n)!=EOF)    {        if(n+m<=0) break;        sum=1;//起点一定算一个答案,如果只有一个点的话，那么答案将输出0        for(int i=0;i<n;i++)        {            scanf("%s",s[i]);            for(int j=0;j<m;j++)            {                if(s[i][j]=='@')//起点已经算了一个，直接修改成#                {                    sx=i;                    sy=j;                    s[i][j]='#';                    break;                }            }        }        dfs(sx,sy);        printf("%d\n",sum);    }    return 0;}