DFS深搜——Red and Black——A Knight's Journey

深搜，从一点向各处搜找到全部能走的地方。

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

45
59
6
13

Source

Asia 2004, Ehime (Japan), Japan Domestic 

代码：

#include<iostream>
using namespace std;
char map[22][22];//定义最大数组
int sum,l,h;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; //四个方位，上、下、左、右
bool border(int x,int y)//推断是否超范围
{
    if(x<0||x>=h||y<0||y>=l) return 0;
    return 1;
}
void search(int x,int y)
{
    int i;
    int xx,yy;
    sum++;//记录长度
    map[x][y]='#';//标记为已走
    for(i=0;i<4;i++) //以当前位置向四个方向扩展
    {
        xx=x+dir[i][0];
        yy=y+dir[i][1];
        if(border(xx,yy)&&map[xx][yy]=='.')  //满足条件就以当前位置继续扩展
        search(xx,yy);
    }
}
int main()
{
    int i,j;
    int x0,y0;
    while(cin>>l>>h)
    {
        sum=0;
        if(l==0&&h==0)break;
        for(i=0;i<h;i++)
        {
            for(j=0;j<l;j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='@')//记录当前位置
                {
                    x0=i;
                    y0=j;
                }
            }
        }
        search(x0,y0);//调用当前位置
        cout<<sum<<endl;
    }
    return 0;
}

A Knight's Journey

Description

Background 
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey 
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans? 

Problem 
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number. 
If no such path exist, you should output impossible on a single line.

Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

八个方向的深搜回溯 把移动方向打好 （网上好多人说要按字典序走才干A   測试了一下  不按字典序也A了）

代码：

#include<iostream>
#include<cstring>
#define M 30
int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
using namespace std;
int cas,n,m,tag;
int map[M][M],t;
char ans[900][2];
void dfs(int x,int y,int k)
{
    int xx,yy,i,j;
    if(k==n*m)
    {
        tag=1;
    }
    for(i=0;i<8;i++)
    {
        xx=x+dx[i];
        yy=y+dy[i];
        if(map[xx][yy]==0&&xx>0&&xx<=n&&yy>0&&yy<=m)
        {
            ans[k][0]=ans[k-1][0]+dy[i];//注意这里x轴移动的位移并非字母轴的位移而是数字轴的位移。。坑我好久
            ans[k][1]=ans[k-1][1]+dx[i];
            map[xx][yy]=1;
for(i=1;i<=n;i++)
{
    for(j=1;j<=m;j++)
    cout<<map[i][j];cout<<endl;
}cout<<endl;
            dfs(xx,yy,k+1);
            if(tag)
                return ;
            map[xx][yy]=0;
        }
    }
    //return ;
}
int main()
{
    int i,j,l=1;
    cin>>cas;
    while(l<=cas)
    {
        memset(map,0,sizeof(map));
        memset(ans,'0',sizeof(ans));
        map[1][1]=1;
        ans[0][0]='A';
        ans[0][1]='1';
        cin>>n>>m;
        t=1;
        tag=0;
        cout<<"Scenario #"<<l<<":"<<endl;
        dfs(1,1,1);
            if(tag)
               {
                   //cout<<ans[0][0]<<ans[0][1];
                   for(i=0;i<n*m;i++)
                    cout<<ans[i][0]<<ans[i][1];
               }
            else
                cout<<"impossible";
            cout<<endl<<endl;
            l++;
    }
}