POJ 1979 Red and Black--DFS

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 29614		Accepted: 16095

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

Test instructions is starting at the @ point and then traversing the entire graph, '. ' Indicate where you can reach, ask for the maximum number of '. ', the initial point is also counted, note sum=1 initialization

#include <stdio.h> #include <string.h> #include <algorithm>using namespace Std;char map[25][25];int Vis[25][25];int n,m,num;int dr[4]={0,1,-1,0};int dc[4]={-1,0,0,1};void dfs (int x,int y) {int i;for (i=0;i<4;i++) {int Dx=x+dr[i];int dy=y+dc[i];if (Dx>=0&&dx<n&&dy>=0&&dy<m&&!vis[dx][dy] &&map[dx][dy]== '. ') {vis[dx][dy]=1;num++;d fs (Dx,dy);}}} int main () {int I,j,x,y;//freopen ("Test.txt", "R", stdin), while (~SCANF ("%d%d", &m,&n)) {if (m==0&&n==0 ) Break;char C[25];for (i=0;i<n;i++) {scanf ("%s", &c), for (j=0;j<m;j++) {map[i][j]=c[j];if (map[i][j]== ' @ ') { X=i;y=j;}}} printf ("%d%d\n", x, y), memset (vis,0,sizeof (Vis)), Num=1;dfs (x, y);p rintf ("%d\n", num);} return 0;}

POJ 1979 Red and Black--DFS