A Knight's Journey

题目链接http://noi.openjudge.cn/ch0205/1490/

描述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.输入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, . . .输出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.样例输入

3
1 1
2 3
4 3

样例输出

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

来源TUD Programming Contest 2005, Darmstadt, Germany

WA:

刚开始思路就错了，我想的是遍历所有的位置，看最后遍历到的总数是不是P*Q。事实上，只要是遍历，最后的结果都是P*Q

这里其实是搜索出一条路径

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#define DEBUG(x) cout<<#x<<" = "<<x<<endl
using namespace std;
int P,Q;
int visited[10][10];
int dirx[]={-1,-2,-2,-1,+1,+2,+2,+1};
int diry[]={-2,-1,+1,+2,-2,-1,+1,+2};
int cnt=0;
void dfs(int i,int j)
{
    if(i<0||j<0||i>=P||j>=Q)return;
    if(visited[i][j])return ;
    visited[i][j]=1;
    cnt++;
    DEBUG(i);
    DEBUG(j);
    printf("\n");
    for(int k=0;k<8 ;k++ ){
        int ii=i+dirx[k];
        int jj=j+diry[k];
        dfs(ii,jj);
    }
}
int main()
{
    freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    int n=1;
    while(t--){
        scanf("%d%d",&P,&Q);
        for(int j=0;j<Q ;j++ ){
            for(int i=0;i<Q ;i++ ){
                memset(visited,0,sizeof(visited));
                cnt=0;
                dfs(i,j);
                DEBUG(i);
                DEBUG(j);
                DEBUG(cnt);
                printf("\n");
            }
        }
    }
}

AC：

刚开始还是不停的WA，搜了一下，发现原因了

题目要求以"lexicographically"方式输出，也就是字典序...要以字典序输出路径，那么搜索的方向（我的程序是path()函数）就要以特殊的顺序排列了...这样只要每次从dfs(A,1)开始搜索，第一个成功遍历的路径一定是以字典序排列...

下图是搜索的次序，马的位置为当前位置，序号格为测试下一步的位置的测试先后顺序

按这个顺序测试，那么第一次成功周游的顺序就是字典序

 

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#define DEBUG(x) cout<<#x<<" = "<<x<<endl
using namespace std;
int P,Q;
int visited[30][30];
//int dirx[]={-1,-2,-2,-1,+1,+2,+2,+1};
//int diry[]={-2,-1,+1,+2,-2,-1,+1,+2};
int dirx[]={-1,+1,-2,+2,-2,+2,-1,+1};
int diry[]={-2,-2,-1,-1,+1,+1,+2,+2};
struct Pos{
    int x;
    int y;
};
Pos path[30];
bool dfs(int i,int j,int step)
{///已经走了step步，从i,j出发，看是否成功
    if(step==P*Q)return true;
    if(i<0||j<0||i>=P||j>=Q)return false;
    if(visited[i][j])return false;
    visited[i][j]=1;
    path[step].x=i;
    path[step].y=j;
    for(int k=0;k<8 ;k++ ){
        int ii=i+dirx[k];
        int jj=j+diry[k];
        if(dfs(ii,jj,step+1))return true;
    }
    visited[i][j]=0;
    return false;
}
int main()
{
//    freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    int n=1;
    while(t--){
        scanf("%d%d",&P,&Q);///P代表数字，Q代表字母，P行Q列
        memset(visited,0,sizeof(visited));
        bool f=false;
        for(int j=0;j<Q ;j++ ){
            for(int i=0;i<P ;i++ ){
                if(dfs(i,j,0)){
                    f=true;
                    break;
                }
            }
            if(f)break;
        }
        printf("Scenario #%d:\n",n++);
        if(f){
           for(int i=0;i<P*Q ;i++ ){
                printf("%c%d",path[i].y+'A',path[i].x+1);
           }
        }
        else printf("impossible");
        printf("\n\n");
    }
}

参考文献

https://blog.csdn.net/lyy289065406/article/details/6647666