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
题目简单翻译：
从‘@’点出发，问能到达的最多的点有多少，‘#’不可经过，计算结果包括‘@’。
解题思路：
广度优先搜索，直接求出到过多少个点。
代码：
#include<cstdio>
#include<cstring> using namespace std;
int n,m,sx,sy;
char mp[][];
int vis[][];
int dx[]={,,,-};
int dy[]={,-,,};
bool check(int x,int y)
{
return x>=&&x<n&&y>=&&y<m;
}
struct node
{
int x,y;
}St[]; int bfs()//手写的队列，队列的尾端就是到达的点的数量
{
memset(vis,,sizeof vis);
int st=,en=;
vis[St[].x][St[].y]=;
while(st<en)
{
node e=St[st++];
for(int i=;i<;i++)
{
node w=e;
w.x=e.x+dx[i],w.y=e.y+dy[i];
if(check(w.x,w.y)&&vis[w.x][w.y]==&&mp[w.x][w.y]!='#')
{
vis[w.x][w.y]=;
St[en++]=w;
}
}
}
return en;
} int main()
{
while(scanf("%d%d",&m,&n)!=EOF&&(n||m))
{
for(int i=;i<n;i++)
{
scanf("%s",mp[i]);
for(int j=;j<m;j++)
if(mp[i][j]=='@')
St[].x=i,St[].y=j;
}
printf("%d\n",bfs());
}
return ;
}
Red and Black