A Knight's Journey POJ

题目链接:点这里

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

题意:

一个骑士要走完棋盘上所有的点一次且仅一次,起点和终点不限,求走的顺序并输出对应的下标,而且尽可能的优先输出字典序最小的那个.

思路:

典型的搜索,只是每次搜索的顺序要使字典序最小的顺序来搜索,然后用数组记录路径就OK了.

代码:

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#include<iostream>using namespace std;int n,m;int path[67];char pat[67];bool vis[65][65];bool check;int dx[]={-2,-2,-1,-1,+1,+1,2,2}; // 骑士走的位置按字典序最小的来排列int dy[]={-1,1,-2,+2,-2,+2,-1,1};void dfs(int x,int y,int k){    if(k==m*n)        check=true;    if(check)        return ;    for(int i=0;i<8;++i){        int sx=x+dx[i];        int sy=y+dy[i];        if(sx>0&&sy>0&&sx<=n&&sy<=m&&vis[sx][sy]==false){            vis[sx][sy]=true;            path[k+1]=sy;            pat[k+1]='A'+sx-1;            dfs(sx,sy,k+1);            vis[sx][sy]=false;            if(check)                return ;        }    }}int main(){    int cas=0;    int t;    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&m);        swap(n,m);        memset(vis,false,sizeof(vis));        check=false;        for(int i=1;i<=n;++i){        for(int j=1;j<=m;++j){            path[1]=j;            pat[1]='A'+i-1;            vis[i][j]=true;            dfs(i,j,1);//从某一点开始搜索            vis[i][j]=false;            if(check)//如果搜到就输出解                break;        }        if(check)            break;        }        cout<<"Scenario #"<<++cas<<':'<<endl;        if(check)        for(int i=1;i<=n*m;++i)            cout<<pat[i]<<path[i];        else cout<<"impossible";        cout<<endl;    if(t)        cout<<endl;    }    return 0;}