20170731

    在学校做题第一天，很无奈，只是看完了第一题就登不上去了，第一题是很经典的那道跳马问题，深搜回溯，写得挺快的，可是太慢了，运行个数据要好久，改了好久才改好了，就是对已经跳过的位置进行标记和回溯上出了错。这道题还有一个很重要的地方，叫“字典顺序”，在马的方向上需要特别的排列，这个之前还真没注意到。

  题目：

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

代码如下：

#include<iostream>
#include<cstring>
using namespace std;
int p,q,sum,flag;
int vis[100][100];
int road[10000][3];
/*
int dx[8]={1,1,2,2,-1,-1,-2,-2};
int dy[8]={2,-2,1,-1,2,-2,1,-1};
*/

int dx[10]={-1,1,-2,2,-2,2,-1,1};//改成字典顺序
int dy[10]={-2,-2,-1,-1,1,1,2,2};


bool ok(int x,int y)
{
    if(x>0&&x<=p&&y>0&&y<=q)
        return 1;
    return 0;
}
void ioan(int x,int y,int t)
{
    road[t][1]=x;
    road[t][2]=y;
    vis[x][y]=1;
    if(t==sum)
    {
        flag=1;
        return ;
    }
    for(int k=0;k<8;k++)
    {
        int xx=x+dx[k];
        int yy=y+dy[k];
        if(ok(xx,yy))
        {
            if(vis[xx][yy]==0)
            {
                vis[xx][yy]=1;
                ioan(xx,yy,t+1);
                vis[xx][yy]=0;
                if(flag==1)
                    return ;
            }
        }
    }
}
int main()
{
    int n;
    int cnt=1;
    cin>>n;
    while(n--)
    {
        cin>>p>>q;
        sum=p*q;
        flag=0;
        memset(vis,0,sizeof(vis));

        for(int i=1;i<=p;i++)
        {
            for(int j=1;j<=q;j++)
            {
                vis[i][j]=1;
                ioan(i,j,1);
                vis[i][j]=0;
                if(flag==1)
                    break;
            }
            if(flag==1)
                break;
        }
        cout<<"Scenario #"<<cnt<<":"<<endl;
        cnt++;
        if(flag==1)
        {
            for(int i=1;i<=sum;i++)
            {
                cout<<char(road[i][2]-1+'A');
                cout<<road[i][1];
            }
            cout<<endl;
        }
        else
            cout<<"impossible"<<endl;
        if(n)
            cout<<endl;
    }

    return 0;
}

    之后下午也没有在做题，看了看树和图论的一些知识和简单题，感觉这里还是个难点，这些天得重点做一下才行。