• poj2488


                                                                                      A Knight's Journey
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 25463   Accepted: 8672

    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




    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<string>
    #include<algorithm>
    using namespace std;
    int dir[8][2]={-1,-2, 1,-2, -2,-1, 2,-1, -2,1, 2,1, -1,2, 1,2};
    bool flag ;
    int p,q,i;    int n;
    const int maxn = 30;
    int sign[30][30];
    struct node 
    {
        int x,y;
    }path[maxn];
    void dfs(int x,int y,int step)
    {
        sign[x][y] = 1;
        path[step].x = x;path[step].y = y;
        if(step==p*q-1)
        {
            flag = 1;
            for(i=0;i<=step;i++)
            {
                printf("%c%d",path[i].y+'A',path[i].x+1);
            }
            printf("
    
    ");
        }
    //    printf("%c%d
    ",path[step].y+'A',path[step].x+1);
        for(i=0;i<8;i++)
        {
            int xx = x + dir[i][0];
            int yy = y + dir[i][1];
            if(xx>=p||yy>=q||xx<0||yy<0)
                continue; 
            if(!sign[xx][yy])
            {
                sign[xx][yy] = 1;
                dfs(xx,yy,step+1);
                sign[xx][yy] = 0;
            }
        }
    }
    int main()
    {
        
        scanf("%d",&n);
        int h =1;
        while(n--)
        {
            flag = 0;
            scanf("%d %d",&p,&q);
            memset(sign,0,sizeof(sign));
            printf("Scenario #%d:
    ",h++);
            dfs(0,0,0);
            if(!flag)
            {
                printf("impossible
    
    ");
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/3221615.html
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