Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25678 Accepted Submission(s): 15498
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.
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)
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
代码：
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
int h,w,t;
int mp[21][21];
int dir[2][4]={ {1,-1,0,0},{0,0,1,-1} };
char s[21][21];
struct node
{
int x,y;
};
queue<node> que;
void BFS(int x,int y)
{
node u,v;
mp[x][y]=1;
u.x=x;
u.y=y;
que.push(u);
while(!que.empty())
{
v=que.front();
que.pop();
for(int i=0;i<4;i++)
{
u.x=v.x+dir[0][i];
u.y=v.y+dir[1][i];
if(s[u.x][u.y]=='.' && !mp[u.x][u.y])
{
//BFS(u.x,u.y);
mp[u.x][u.y]=1;
que.push(u);
t++;
}
}
}
}
int main()
{
int i,j;
while(cin>>w>>h && h,w)
{
t=1;
memset(s,0,sizeof(s));
memset(mp,0,sizeof(mp));
for(i=0;i<h;i++)
for(j=0;j<w;j++)
cin>>s[i][j];
for(i=0;i<h;i++)
{
for(j=0;j<w;j++)
if(s[i][j]=='@') break;
if(s[i][j]=='@') break;
}
BFS(i,j);
cout<<t<<endl;
}
return 0;
}