Red and Black
Time Limit: 1000MS |
|
Memory Limit: 30000K |
Total Submissions: 25797 |
|
Accepted: 13967 |
Description There is a rectangular and covered with square tiles. Each tile is colored either red or black. A man was standing on a black tile. From a tiles, he can move to one of the four adjacent tiles. But he can ' t move on red tiles, and 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 the positive integers W and H; W and H is the numbers of tiles in the X-and y-directions, respectively. W and H is not more than 20.
There is H more lines in the data set, and each of the 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 the 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.#..........#.#######. #.#.....#.. #.#.###.#.. #.#[email protected]#.#. #.#####.#.. #.......#.. #########............ 11 6..#. #.. #....#.. #.. #....#.. #.. ###.. #.. #.. #@...#.. #.. #....#.. #.. #.. 7 7..#.#....#.#. ###.###[email protected]###.###. #.#....#.#.. 0 0
Sample Output 4559613
Source Japan 2004 Domestic |
[Submit] [Go back] [Status] [Discuss]
Just attach the code. It's a simple, deep search.
#include <stdio.h> #include <string.h>char map[25][25];int dir[4][2]={1,0,-1,0,0,1,0,-1},sum,n,m;bool Limit (int x,int y)//Whether out of bounds {if (x>=0&&x<n&&y>=0&&y<m) return True;elsereturn false;} void Dfs (int x,int y) {map[x][y]= ' # ';//update floor for (int i=0;i<4;i++) {int xx=x+dir[i][0];int yy=y+dir[i][1];if (limit (XX, yy) &&map[xx][yy]== '. ') {sum++;d fs (XX,YY);}}} int main () {while (scanf ("%d%d", &m,&n)!=eof) {if (n==0&&m==0) Break;memset (map,0,sizeof (map)); int i=0;i<n;i++) scanf ("%s", Map[i]), sum=0;for (int i=0;i<n;i++) for (int j=0;j<m;j++) if (map[i][j]== ' @ ') {DFS ( I,J); break;} printf ("%d\n", sum+1);//Starting point is also, so +1}return 0;}
Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.
poj1979 Red and Black (DFS)