POJ1979 Red and Black (DFS)

       搜索听学长们是经常谈论的，由于一开始都是在做很水很水的题目，这种类型的题倒也是初次接触。此题还是比较简单的，利于我们这种初学者上手。

题目的大意是说给你一个N*M的字符矩阵，‘#’为墙不可走、‘@’为起点。问你现在从这一点我们遍历全图、在不走回头路的前提下最多走多少步。

题目意思一明白实现就简单多了（英语实在是差，能看实例分析出题意的题实在是少啊。。。）、用一个V[ ]数组来判断某点是否走过、避免走回头路。

在处理的过程中还要要注意的便是边界的判断，如果是初次做很容易就会超界。

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 26238 Accepted: 14249

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) 
The end of the input is indicated by a line consisting of two 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.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

Sample Output

4559613

Source

Japan 2004 Domestic

#include<iostream>#include<cstring>#include<string>#define N 25using namespace std;int w,h,startx,starty,count;int v[N][N];char arr[N][N];int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};//定义某点可走的四个方向 void DFS(int x,int y){for(int i=0;i<4;i++){int mx=x+dir[i][0];int my=y+dir[i][1];if(arr[mx][my]=='.'&&v[mx][my]==0&&mx>=0&&mx<w&&my>=0&&my<h){v[mx][my]=1;count++;DFS(mx,my);}}}int main(){int i,j;while(cin>>h>>w,w||h){memset(v,0,sizeof(v));for(i=0;i<w;i++)  for(j=0;j<h;j++)   {       cin>>arr[i][j];       if(arr[i][j]=='@')       {       startx=i;       starty=j;   }} count=1;v[startx][starty]=1;DFS(startx,starty);cout<<count<<endl;}return 0;}