Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6453 Accepted Submission(s): 4081
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.
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
Source
Asia 2004, Ehime (Japan), Japan Domestic
Recommend
Eddy
題意:紅格子和黑格子。人只能走黑格子。數人能走多少個黑格子並輸出。
代碼:
#include<cstdio> #include<iostream> #include<cstring> using namespace std; char maps[25][25]; int r,c,sx,sy,res; int dx[]={-1,1,0,0}; int dy[]={0,0,-1,1}; void dfs(int a,int b) { for(int i=0;i<4;i++) { int xx=a+dx[i]; int yy=b+dy[i]; if(xx>=1&&xx<=r&&yy>=1&&yy<=c&&maps[xx][yy]=='.') { maps[xx][yy]='#'; res++; dfs(xx,yy); } } } int main() { int i,j; while(scanf("%d%d",&c,&r)&&r&&c) { memset(maps,0,sizeof(maps)); res=1; getchar(); for(i=1;i<=r;i++) { for(j=1;j<=c;j++) { scanf("%c",&maps[i][j]); if(maps[i][j]=='@') { sx=i; sy=j; maps[i][j]='#'; } } getchar(); } dfs(sx,sy); printf("%d\n",res); } return 0; }