A Knight's Journey 分类： POJ 搜索 2015-08-08 07:32 2人阅读 评论(0) 收藏

A Knight’s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 35564 Accepted: 12119

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

一段时间没有做搜索,果然手生了很多,犯了各种各样的错误,要赶紧补回来.

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker,"/STACK:102400000")
#define WW freopen("output.txt","w",stdout)

struct node
{
    int x;
    int y;
} a[100];

int Dir[][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};

bool vis[9][9];

int n,m;

int flag;

bool DFS(int x,int y,int site)
{
    vis[x][y]=true;
    a[site].x=x;
    a[site].y=y;
    if(site==n*m)
    {
        flag=true;
        return true;
    }
    int Fx,Fy;
    for(int i=0; i<8; i++)
    {
        Fx=x+Dir[i][0];
        Fy=y+Dir[i][1];
        if(Fx>=0&&Fx<n&&Fy>=0&&Fy<m&&!vis[Fx][Fy])
        {
            if(DFS(Fx,Fy,site+1))
            {
                return true;
            }

        }
    }
    vis[x][y]=false;
    return false;
}

int main()
{

    int T,w=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        flag=false;
        memset(vis,false,sizeof(vis));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(DFS(i,j,1))
                {
                    flag=true;
                    break;
                }
            }
            if(flag)
            {
                break;
            }
        }
        printf("Scenario #%d:\n",w++);
        if(flag)
        {
            for(int i=1; i<=n*m; i++)
            {
                printf("%c%c",a[i].y+'A',a[i].x+1+'0');
            }
            printf("\n\n");
        }
        else
        {
            printf("impossible\n\n");
        }
    }
    return 0;
}

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