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
题目链接:http://poj.org/problem?id=2488
解法类型:回溯算法
解题思路:这是一道经典的回溯题,直接利用系统栈深度优先搜索即可。用一个数组标记走过的路,如果无路可走就取消标记,一直搜索下去。如果全部被标记,即搜索成功,未全部被标记则搜索失败。但需要注意输出是按字典序的顺序输出,所以要从A1出发搜索。
算法实现:
//STATUS:C++_AC_16MS_168K
#include
#include
const int MAXN=10;
int DFS(int tot,int rear,int x,int y);
int p,q,way[MAXN*MAXN][2]={0},vis[MAXN][MAXN];
int dx[8]={-2,-2,-1,-1,1,1,2,2}, //按字典序方向行走
dy[8]={-1,1,-2,2,-2,2,-1,1};
int main()
{
// freopen("in.txt","r",stdin);
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
memset(vis,0,sizeof(vis)); //预处理
vis[0][0]=1; //标记A1已经经过
scanf("%d%d",&p,&q);
printf("Scenario #%d:\n",i);
if(DFS(p*q,1,0,0)){
for(int j=0,tot=p*q;j=0&&nx=0&&ny
PS:今天刚从医院出来,已经有十多天没有碰过代码了,昨天下午小试身手参加了中南的月赛,结果是大跌眼镜啊,只A掉了两个。有些题在提交时因忘了注销freopen而不知道多提交了几次,可怜我在比赛时还一直郁闷怎么老是WA。。T.T...还有一道题,死磕了两小时,最后才发现题目居然看错了,结果直接悲剧。
翻了翻以前的题解,发现一般是直接贴题目上代码,以后就不能这么随便了,解题报告一定要写得规范,那就从今天开始吧。为此,我的解题报告格式为:
题目描述
解法类型
解题思路
算法实现