poj2488

A Knight's Journey Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14310 Accepted: 4766 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

 

DFS

#include 
#include 

using namespace std;

typedef struct
{
    int p;
    int q;
}Node;

bool map[27][27];
Node record[900];
int p,q;
bool isOK;
const int xx[8]={-1, 1, -2, 2, -2, 2, -1, 1};
const int yy[8]={-2, -2, -1, -1, 1, 1, 2, 2};

bool check(Node temp)
{
    return temp.p<=p && temp.q<=q && temp.p>=1 && temp.q>=1 && !map[temp.p][temp.q];
}

void DFS(int x, int y, int step)
{
    Node temp;

    if(step==p*q+1)
    {
        isOK=true;
        for(int i=1; i<=p*q; i++)
        {
            cout<>case_no;
    for(int cas=1; cas<=case_no; cas++)
    {
        cout<<"Scenario #"<>p>>q;
        step=1;

        isOK=false;
        map[1][1]=true;
        record[step].p=1;
        record[step].q=1;

        DFS(1,1,step+1);

        if(!isOK)
            cout<<"impossible";
        cout<

转载于:https://www.cnblogs.com/eric-blog/archive/2011/05/04/2036912.html