POJ2488A Knight's Journey[DFS]

Posted 2020-08-11 Candy?

A Knight‘s Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41936		Accepted: 14269

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

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//
//  main.cpp
//  poj2488
//
//  Created by Candy on 9/27/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=50;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x;
}
int T,n,m,sum,vis[N][N],flag=0,cas=0;
struct data{
    int x,y;
    data(int a=0,int b=0):x(a),y(b){}
}path[N];
int dx[8]={-1,1,-2,2,-2,2,-1,1},dy[8]={-2,-2,-1,-1,1,1,2,2};
void print(){
    for(int i=1;i<=sum;i++){
        int x=path[i].x,y=path[i].y;
        printf("%c%d",‘A‘-1+y,x);
    }
}
void dfs(int x,int y,int d){//printf("dfs %d %d %d\n",x,y,d);
    path[d]=data(x,y);
    if(d==sum){flag=1;return;}
    for(int i=0;i<8;i++){
        int nx=x+dx[i],ny=y+dy[i];
        if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!vis[nx][ny]&&!flag){
            vis[nx][ny]=1;
            dfs(nx,ny,d+1);
            vis[nx][ny]=0;
        }
    }
}

int main(int argc, const char * argv[]) {
    T=read();
    while(T--){
        n=read();m=read();
        sum=n*m; flag=0;
        memset(vis,0,sizeof(vis));
        vis[1][1]=1;
        dfs(1,1,1);
        printf("Scenario #%d:\n",++cas);
        if(!flag) printf("impossible");else print();
        printf("\n\n");
    }
    return 0;
}