HDOJ 1312 Red and Black （简单dfs）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12655    Accepted Submission(s): 7810

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

45596ac代码：
#include<stdio.h>#include<string.h>char map[25][25];//int v[25][25];int n,m,num;int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};int check(int xx,int yy){if(xx<0||xx>=n||yy<0||yy>=m||map[xx][yy]=='#')return 1;return 0;}void dfs(int x,int y){int i;map[x][y]='#';num++;for(i=0;i<4;i++){int nx=x+dir[i][0];int ny=y+dir[i][1];if(check(nx,ny))continue;dfs(nx,ny);}}int main(){int bx,by,i,j;while(scanf("%d%d",&m,&n)!=EOF){//memset(v,0,sizeof(v));if(n==0&&m==0)break;for(i=0;i<n;i++){    scanf("%s",map[i]);    for(j=0;j<m;j++)    {    if(map[i][j]=='@')    {    bx=i;    by=j;}}}num=0;dfs(bx,by);printf("%d\n",num);}return 0;}