hdu1312Red and Black （dfs）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16348    Accepted Submission(s): 10071

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

4559613

题意：给你一个图，#代表墙， . 代表通路，求出能走最多能走多少个格子

思路：dfs基础题，水题。

Code：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cstdlib>#include <cmath>#include <algorithm>using namespace std;const int maxn = 105;char map[25][25];int vis[25][25],cnt,n,m,dir[4][2]={0,1,0,-1,1,0,-1,0};void dfs(int x,int y){    if(x<0||x>=n||y<0||y>=m||vis[x][y]||map[x][y]=='#')        return ;    int fx,fy,i;    cnt++;    vis[x][y]=1;    for(i=0;i<4;i++)    {        fx=x+dir[i][0];        fy=y+dir[i][1];        dfs(fx,fy);    }}int main(){    int sx,sy,i,j;    while(scanf("%d%d",&m,&n))    {        getchar();        if(m==0&&n==0)            break;        memset(vis,0,sizeof(vis));        for(i=0;i<n;i++){             for(j=0;j<m;j++){              scanf("%c",&map[i][j]);              if(map[i][j]=='@')              {                  sx=i;                  sy=j;              }           }           getchar();        }        cnt=0;        dfs(sx,sy);        printf("%d\n",cnt);    }    return 0;}