NOI2.5 1490:A Knight's Journey

描述

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.

输入

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, . . .

输出

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.

样例输入

31 12 34 3

样例输出

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

题目大意：给定棋盘长和宽，一马从左上角开始，问能否遍历棋盘，如果能，输出字典序最小的路径，如果不能，输出“impossible”

这道题很像马走日（我的博客里有），只是要输出路径，还要字典序最小的，所以要注意这匹马优先选择的走法，其他没什么难度

代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int k,m,n,w[8]={-1,1,-2,2,-2,2,-1,1},u[8]={-2,-2,-1,-1,1,1,2,2};//注意这里的优先选择路径int a[1001][2];int v[100][100];bool check(int x,int y){if(x>=1&&x<=m&&y>=1&&y<=n&&!v[x][y])return 1;return 0;}void find(int x,int y,int s){int i;if(k==0){a[s][0]=x;a[s][1]=y;if(s==m*n){k=1;return;}}for(i=0;i<8;i++)if(check(x+w[i],y+u[i])){v[x][y]=1;find(x+w[i],y+u[i],s+1);v[x][y]=0;}}int main(){int i,j,p;scanf("%d",&p);for(i=0;i<p;i++){k=0;scanf("%d%d",&m,&n);find(1,1,1);printf("Scenario #%d:\n",i+1);if(k==0)printf("impossible\n\n");else{for(j=1;j<=m*n;j++)printf("%c%d",a[j][1]+64,a[j][0]);printf("\n\n");}}}

难度不大，注意细节