• 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;
    }
  • 相关阅读:
    【codecombat】 试玩全攻略 第二章 边远地区的森林 一步错
    【codecombat】 试玩全攻略 第十八关 最后的kithman族
    【codecombat】 试玩全攻略 第二章 边远地区的森林 woodlang cubbies
    【codecombat】 试玩全攻略 第二章 边远地区的森林 羊肠小道
    【codecombat】 试玩全攻略 第十七关 混乱的梦境
    【codecombat】 试玩全攻略 第二章 边远地区的森林 林中的死亡回避
    【codecombat】 试玩全攻略 特别关:kithguard斗殴
    【codecombat】 试玩全攻略 第二章 边远地区的森林 森林保卫战
    【codecombat】 试玩全攻略 第二章 边远地区的森林
    实验3 类和对象||
  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/3221615.html
Copyright © 2020-2023  润新知