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poj 2488

2024-07-13 22:58:06   221人阅读

A Knight‘s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 30099 Accepted: 10320

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

AC代码：

#include<iostream>
#include<cstring>
using namespace std;
int p,q;
int G[30][30];
int sucess;
char ch[1000];
int a[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int dfs(int x,int y,int step){
    if(x<0 || x>p-1 || y<0 || y>q-1 || G[x][y]) return 0;
    G[x][y]=1;
    if(step==p*q){
        ch[2*step]='\0';
        return 1;
    }
    for(int k=0;k<8;k++){
        if(dfs(x+a[k][0],y+a[k][1],step+1)){
            ch[2*(step+1)-2]=y+a[k][1]+'A';
            ch[2*(step+1)-1]=x+a[k][0]+'1';
            return 1;
        }
    }
    G[x][y]=0;
    return 0;
}
int main(){
    int T; cin>>T;
    for(int cas=1;cas<=T;cas++){
        cin>>p>>q;
        for(int j=0;j<q;j++){
            for(int i=0;i<p;i++){
                sucess=0;
                memset(G,0,sizeof(G));
                ch[0]=j+'A';
                ch[1]=i+'1';
                if(dfs(i,j,1)){
                    sucess=1;
                    break;
                }
            }
            if(sucess)
                break;
        }
        cout<<"Scenario #"<<cas<<":"<<endl;
        if(!sucess)
            cout<<"impossible"<<endl<<endl;
        else
            cout<<ch<<endl<<endl;
    }
    return 0;
}



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