Problem P：Red and Black（HDU 1312）

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
    .#.........
    .#.#######.
    .#.#.....
4000
#.
    .#.#.###.#.
    .#.#..@#.#.
    .#.#####.#.
    .#.......#.
    .#########.
    ...........
    11 6
    ..#..#..#..
    ..#..#..#..
    ..#..#..###
    ..#..#..#@.
    ..#..#..#..
    ..#..#..#..
    7 7
    ..#.#..
    ..#.#..
    ###.###
    ...@...
    ###.###
    ..#.#..
    ..#.#..
    0 0

Sample Output

   45
   59
   6
   13————————————————————————————————————————————思路：经典搜索，bfs、dfs均可，注意输入房间长宽时，先输入列数再输入行数

Source Program

#include<cstdio>
int w,h;//w是列，h是行
int sum;
char map[21][21];
int direction[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//方向数组

struct node
{
int x;
int y;
}quene[500];

bool judge(int x0,int y0)//判断坐标是否在规定范围内
{
if(x0>=0&&x0<h&&y0>=0&&y0<w)
return true;
else
return false;
}

void bfs(int x0,int y0)
{
int head,tail;
int x,y;
int i;

head=0,tail=1;//设置队列首尾初值
quene[1].x=x0;quene[1].y=y0;//初始状态存入队列
while(head<tail)
{
head++;//队首加1，出队
for(i=0;i<4;i++)//依次向上下左右搜索
{
x=quene[head].x+direction[i][0];
y=quene[head].y+direction[i][1];
if(map[x][y]=='.'&&judge(x,y))//若子节点符合条件
{
map[x][y]='#';//若走过，标记，不能重复走

quene[++tail].x=x;//入队
quene[tail].y=y;//入队
sum++;//走过数+1
}
}
}
}
int main()
{
int x0,y0;
int i,j;

while(scanf("%d %d",&w,&h)!=EOF)//输入房间长与宽
{
getchar();
if(w==0&&h==0)  break;//大小为0时终止循环

for(i=0;i<h;i++)
{
for(j=0;j<w;j++)
{
scanf("%c",&map[i][j]);//输入每块瓷砖
if(map[i][j]=='@')//记录起始位置
{
x0=i;
y0=j;
}
}
getchar();
}
sum=1;
bfs(x0,y0);//搜索
printf("%d\n",sum);
}
return 0;
}