poj1979Red and Black(广度优先搜索)

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 14020		Accepted: 7218

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

45
59
6
13

广搜的一道基础题，但是要注意程序的组织结构

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int visited[20][20];
char map[20][20];
struct point{
int x,y;
};
int move[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int w,h;
int BFS(point s)
{
int i;
int cx,cy;
queue<point> qu;
visited[s.x][s.y]=1;
int max=1;
qu.push(s);
while(!qu.empty())
{
point temp=qu.front();
qu.pop();
for(i=0;i<4;i++)
{
cx=temp.x+move[i][0];cy=temp.y+move[i][1];
if(cx>=0&&cx<h&&cy>=0&&cy<w&&!visited[cx][cy]&&map[cx][cy]!='#')
{
max++;
visited[cx][cy]=1;
point temp1;
temp1.x=cx;temp1.y=cy;
qu.push(temp1);
}
}
}
return max;
}
int main()
{
int i,j;
while(scanf("%d%d",&w,&h))
{
getchar();
int sx,sy;
point start;
memset(visited,0,sizeof(visited));
if(w==0&&h==0) break;
for(i=0;i<h;i++)
{
for(j=0;j<w;j++)
{
scanf("%c",&map[i][j]);
if(map[i][j]=='@'){sx=i;sy=j;}
}
getchar();
}
start.x=sx;start.y=sy;
int max=BFS(start);
printf("%d/n",max);
}
return 0;
}