Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6905 Accepted Submission(s): 4384
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
Source
Recommend
Eddy
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<deque> 5 #include<iostream> 6 #define maxn 25 7 using namespace std; 8 char maze[maxn][maxn]; 9 int w,h; 10 /*建立方向树*/ 11 struct node 12 { 13 int x,y; 14 }start; 15 /*搜索方向*/ 16 int dir[4][2]= 17 { 18 {0,1}, 19 {0,-1}, 20 {-1,0}, 21 {1,0} 22 }; 23 void bfs() 24 { 25 /*入队，出队用*/ 26 deque<node>q; 27 /*暂存位置*/ 28 node q1,q2; 29 q.push_back(start); 30 while(!q.empty()) 31 { 32 q1=q.front (); 33 q.pop_front(); 34 for(int i=0;i<4;i++) 35 { 36 q2.x=q1.x+dir[i][0]; 37 q2.y=q1.y+dir[i][1]; 38 if(q2.x>h||q2.x<1||q2.y>w||q2.y<1||maze[q2.x][q2.y]=='#'||maze[q2.x][q2.y]=='p') /*return ;*/ 39 ; /*啥也不干，就这么一个逗号....*/ 40 else 41 { 42 if(maze[q2.x][q2.y]=='.') 43 maze[q2.x][q2.y]='p'; /*就暂时用P来代表占位置吧！*/ 44 /*入队*/ 45 q.push_back (q2); 46 } 47 } 48 } 49 } 50 int main() 51 { 52 int i,j,ans; 53 while(scanf("%d%d",&w,&h),h+w) 54 { 55 getchar(); 56 memset(maze,' ',sizeof(maze)); 57 for(i=1;i<=h;i++) 58 { 59 for(j=1;j<=w;j++) 60 { 61 scanf("%c",&maze[i][j]); 62 if(maze[i][j]=='@') 63 { 64 start.x=i; 65 start.y=j; 66 maze[i][j]='#'; 67 } 68 } 69 getchar(); 70 } 71 bfs(); 72 ans=1; 73 for(i=1;i<=h;i++) 74 { 75 for(j=1;j<=w;j++) 76 { 77 78 if(maze[i][j]=='p') 79 ans++; 80 } 81 } 82 printf("%d ",ans); 83 } 84 return 0; 85 }
bfs+匹配