POJ - 1979-Red and Black

Red and Black
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d
& %I64u

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

题意：就是让你从起点“@”开始遍历所有能走到的“.”，可以按原路返回，但不再计数。注意：先输入的w是列数，后输入的h是行数。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int w,h,ans,a,b;
char map[25][25];
bool vis[25][25];//辅助数组，走过的记为true，避免重复计算，陷入死循环
int px[4]={-1,0,1,0};
int py[4]={0,1,0,-1};
int dfs(int a,int b){
for(int i=0;i<4;i++){
int x=a+px[i];
int y=b+py[i];//移动后的位置
if(map[x][y]=='.'&&vis[x][y]==false){
vis[x][y]=true;
ans++;//计数器
dfs(x,y);//递归调用自身，遍历完所有情况
}
}
return ans;
}
int main(){
while(scanf("%d%d",&w,&h),w||h){
memset(vis,false,sizeof(vis));//辅助数组初始化
memset(map,'#',sizeof(map));//将图全部初始化为#，防止越界
ans=0;//计数器初始化
for(int i=1;i<=h;i++){//从1开始！！
getchar();//吸收空行键
for(int j=1;j<=w;j++){
scanf("%c",&map[i][j]);
if(map[i][j]=='@'){
a=i,b=j;
vis[i][j]=true;//记录@的位置，即dfs的起点
}
}
}
printf("%d\n",dfs(a,b)+1);//加上@本身的位置
}
return 0;
}