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

31 12 34 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。