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poj2488--A Knight's Journey(dfs，骑士问题)

2024-07-23 06:57:56   238人阅读

A Knight‘s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 31147 Accepted: 10655

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany
给出m行n列的棋盘，问骑士能不能走完所有的点，每个点只能走一次，要求字典序最小输出。
如果能走完，那么一定可以从A1开始走，字典序最小 只要dfsA1开始，看能不能走完，注意搜索的顺序。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int t , n , m , sum ;int mm[30][30] ;int pre[1000] ;int a[8][2] = { {-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1} };int dfs(int x,int y,int temp){    if( temp == sum )        return 1 ;    int flag = 0 , xx , yy , i ;    for(i = 0 ; i < 8 ; i++)    {        xx = x + a[i][0] ;        yy = y + a[i][1] ;        if( xx >= 0 && xx < n && yy >= 0 && yy < m && !mm[xx][yy] )        {            mm[xx][yy] = 1 ;            pre[x*m+y] = xx*m+yy ;            flag = dfs(xx,yy,temp+1);            if( flag ) return flag ;            mm[xx][yy] = 0 ;        }    }    return flag ;}int main(){    int i , j , k , tt ;    scanf("%d", &t);    for(tt = 1 ; tt <= t ; tt++)    {        memset(mm,0,sizeof(mm));        memset(pre,-1,sizeof(pre));        scanf("%d %d", &m, &n);        sum = n*m ;        mm[0][0] = 1 ;        k = dfs(0,0,1);        printf("Scenario #%d:\n", tt);        if(k == 0)            printf("impossible");        else        {            for(i = 0 , k = 0 ; i < sum ; i++)            {                printf("%c%c", k/m+'A', k%m+'1');                k = pre[k] ;            }        }        printf("\n\n");    }    return 0;}

poj2488--A Knight's Journey(dfs，骑士问题)


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