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
45
59
6
13
Source
Asia 2004, Ehime (Japan), Japan Domestic
思路
平平无奇的一道简单bfs问题，只要每次广搜入队的时候都统计一次就好了，最后返回结果并输出
代码
#include<bits/stdc++.h>
using namespace std;
int a,b,c,t;
const int d[][2]={ {-1,0},{0,1},{1,0},{0,-1} };
struct node
{
int x;
int y;
}st,ed;
int n,m;
char maps[21][21];
bool judge(node x)
{
if(x.x<=m && x.x>=1 && x.y<=n && x.y>=1 && maps[x.x][x.y]=='.')
return true;
return false;
}
int bfs(node st)
{
queue<node> q;
q.push(st);
maps[st.x][st.y] = '#';
node now,next;
int t = 0;
while(!q.empty())
{
now = q.front();
q.pop();
for(int i=0;i<4;i++)
{
next.x = now.x + d[i][0];
next.y = now.y + d[i][1];
if(judge(next))
{
q.push(next);
t++;
maps[next.x][next.y] = '#';
}
}
}
return t+1;//起点也算
}
int main()
{
while(cin>>n>>m)
{
if(n==0 && m==0) break;
for(int i=1;i<=m;i++)
for(int j=1;j<=n;j++)
{
cin >> maps[i][j];
if(maps[i][j]=='@')
{
st.x = i; st.y = j;
}
}
int ans = bfs(st);
cout << ans << endl;
}
return 0;
}