【POJ 2488】A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 38205		Accepted: 12970

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

题目大意：求能以国际象棋中马的规则走完q*p的棋盘中所有格子的所有解中字典序最小的一组。

思路：本题有两个需要注意的地方：1.要求字典序最小的一组解，所以要以特定的顺序去移动棋子，代码中有。2.由于解是要经过棋盘中的所有格子，所以以A1为起点的解一定为所求，所以只对(0,0)点dfs即可。

代码：

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
int p,q;
bool flag,f;
int vis[30][30];
int movement[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
                                                ///因为要按照字典序排列的第一组解，所以移动顺序只能是这样的。
struct Point
{
    int x0;
    char y0;
} ch[900];                                      ///用结构体保存路径
bool check(int x,int y)                         ///判断当前格子是否合法
{
    if(x>0&&x<=p&&y>0&&y<=q&&vis[x][y]==0)
        return true;
    return false;
}
void dfs(int x,int y,int val)
{
    //cout<<x<<" "<<y<<" "<<val<<endl;
    vis[x][y]=1;
    ch[val].x0=x;
    ch[val].y0=y+'A'-1;
    if(val==p*q)
    {
        flag=true;
        return;
    }
    for(int i=0; i<8; i++)
    {
        int tempx=x+movement[i][0];
        int tempy=y+movement[i][1];
        if(check(tempx,tempy)&&!flag)
        {
            dfs(tempx,tempy,val+1);
            vis[tempx][tempy]=0;
        }
    }
}
int main()
{
    int t;
    int cnt=1;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            flag=false;
            scanf("%d%d",&p,&q);
            memset(ch,0,sizeof(ch));
            memset(vis,0,sizeof(vis));
            dfs(1,1,1);
            printf("Scenario #%d:
",cnt++);
            if(!flag)
                printf("impossible
");
            else
            {
                for(int i=1; i<=p*q; i++)
                    printf("%c%d",ch[i].y0,ch[i].x0);
                puts("");
            }
            if(t!=0)
                puts("");
        }
    }
    return 0;
}