poj 2488:A Knight's Journey

总时间限制: 

1000ms 

内存限制: 

65536kB

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

思路：这是一个简单的搜索题，马是按国际象棋的走法，即“两横一竖”或“两竖一横”，则马一共有八种走法，在代码中用move函数表示。move的走法，按照题中给出的字典顺序进行搜索，即b-2，b-1……。只要按字典顺序搜索，则找到的第一条路径就是符合题目要求的输出，不需要按照回溯方法遍历整个图。递归时，首先判断是否已经找到路径，再判断该路径是否已经走过，然后判断马是否走出边界。然后根据上未走的格子数判断马是否已经走完全部格子，若全部走完，则输出返回，否则将走过的该格子填入vector中，并进行相应处理，然后继续递归，最后根据回溯的方法，返回操作。

注意：

1.这里要注意输出是需要按字典顺序的，即不需要遍历搜索整个图，只需要找到按字典顺序输出的第一条记录即可，这就需要安排好搜索的路径，否则还要对输出进行排序，增加时间。

2.进行数组操作的时候注意不要产生runtime error。

代码如下：

#include <iostream>#include <vector>#include <string>#include <stdio.h>using namespace std;char qletter[27]={'0','A','B','C','D','E','F','G','H','I','J',<span style="white-space:pre"></span>'K','L','M','N','O','P','Q','R','S','T','U',<span style="white-space:pre"></span>'V','W','X','Y','Z'};int CHESS[27][27];int totalN,p,q;char buf[10];vector<string> result;bool isfind;void clear(int ix, int jy){<span style="white-space:pre"></span>++ix;<span style="white-space:pre"></span>++jy;<span style="white-space:pre"></span>for (int i = 0; i < ix; ++i)<span style="white-space:pre"></span>for(int j = 0; j < jy; ++j)<span style="white-space:pre"></span>CHESS[i][j] = 0;}void move(int a, int b){<span style="white-space:pre"></span>if (isfind)<span style="white-space:pre"></span>return;<span style="white-space:pre"></span>if (CHESS[a][b])<span style="white-space:pre"></span>return;<span style="white-space:pre"></span>if (a < 1 || b < 1 || a > p || b > q)<span style="white-space:pre"></span>return;<span style="white-space:pre"></span>CHESS[a][b] = 1;<span style="white-space:pre"></span>--totalN;<span style="white-space:pre"></span>sprintf(buf, "%c%d", qletter[b], a);<span style="white-space:pre"></span>string string_buf(buf);<span style="white-space:pre"></span>result.push_back(string_buf);<span style="white-space:pre"></span>if (totalN == 0){<span style="white-space:pre"></span>isfind = true;<span style="white-space:pre"></span>for (vector<string>::iterator it = result.begin(); it != result.end(); ++it){<span style="white-space:pre"></span>cout << *it;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>cout << endl;<span style="white-space:pre"></span><span style="white-space:pre"></span>//result.pop_back();<span style="white-space:pre"></span>//CHESS[a][b] = 0;<span style="white-space:pre"></span>//++totalN;<span style="white-space:pre"></span>return;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>move(a-1, b-2);<span style="white-space:pre"></span>move(a+1, b-2);<span style="white-space:pre"></span>move(a-2, b-1);<span style="white-space:pre"></span>move(a+2, b-1);<span style="white-space:pre"></span>move(a-2, b+1);<span style="white-space:pre"></span>move(a+2, b+1);<span style="white-space:pre"></span>move(a-1, b+2);<span style="white-space:pre"></span>move(a+1, b+2);<span style="white-space:pre"></span>result.pop_back();<span style="white-space:pre"></span>CHESS[a][b] = 0;<span style="white-space:pre"></span>++totalN;}void search(int p, int q){<span style="white-space:pre"></span>move(1,1);<span style="white-space:pre"></span>if (isfind == false)<span style="white-space:pre"></span>cout << "impossible" << endl;}int main(){<span style="white-space:pre"></span>int n;<span style="white-space:pre"></span>int num = 1;<span style="white-space:pre"></span>cin >> n;<span style="white-space:pre"></span>while (n--){<span style="white-space:pre"></span>cin >> p >> q;<span style="white-space:pre"></span>isfind = false;<span style="white-space:pre"></span>totalN = p*q;<span style="white-space:pre"></span>clear(p, q);<span style="white-space:pre"></span>cout << "Scenario #"<<num<<":\n";<span style="white-space:pre"></span>search(p, q);<span style="white-space:pre"></span>result.clear();<span style="white-space:pre"></span>++num;<span style="white-space:pre"></span>cout << "\n";<span style="white-space:pre"></span>}<span style="white-space:pre"></span>return 0;}