Red and Black
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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
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：
1 #include<cstdio> 2 #include<cstring> 3 4 using namespace std; 5 int n,m,sx,sy; 6 char mp[24][24]; 7 int vis[24][24]; 8 int dx[]={0,0,1,-1}; 9 int dy[]={1,-1,0,0}; 10 bool check(int x,int y) 11 { 12 return x>=0&&x<n&&y>=0&&y<m; 13 } 14 struct node 15 { 16 int x,y; 17 }St[1000]; 18 19 int bfs()//手写的队列，队列的尾端就是到达的点的数量 20 { 21 memset(vis,0,sizeof vis); 22 int st=0,en=1; 23 vis[St[0].x][St[0].y]=1; 24 while(st<en) 25 { 26 node e=St[st++]; 27 for(int i=0;i<4;i++) 28 { 29 node w=e; 30 w.x=e.x+dx[i],w.y=e.y+dy[i]; 31 if(check(w.x,w.y)&&vis[w.x][w.y]==0&&mp[w.x][w.y]!='#') 32 { 33 vis[w.x][w.y]=1; 34 St[en++]=w; 35 } 36 } 37 } 38 return en; 39 } 40 41 int main() 42 { 43 while(scanf("%d%d",&m,&n)!=EOF&&(n||m)) 44 { 45 for(int i=0;i<n;i++) 46 { 47 scanf("%s",mp[i]); 48 for(int j=0;j<m;j++) 49 if(mp[i][j]=='@') 50 St[0].x=i,St[0].y=j; 51 } 52 printf("%d ",bfs()); 53 } 54 return 0; 55 }