POJ - 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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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

#include<iostream>#include<string>#include<cstring>#include<algorithm>#include<stack>using namespace std;               //很经典的dfs题，需要记录合法路径，用栈。struct node{char c;                        //要求输出路径的字典序最小，这与搜索的方向先后有关。int r;                         //细心，恒心，注意细节就会做出来。};int p, q;int xt, yt;int mark[30][30];bool isillegal(int x, int y) {if (x >= 1 && x <= p &&y >= 1 && y <= q && !mark[x][y]) return true;else return false;}void move(int casen, int x, int y){switch (casen){case 1:{ xt = x - 1; yt = y - 2; break; }case 2:{ xt = x + 1; yt = y - 2; break; }case 3:{ xt = x - 2; yt = y - 1; break; }case 4:{ xt = x + 2; yt = y - 1; break; }case 5:{ xt = x - 2; yt = y + 1; break; }case 6:{ xt = x + 2; yt = y + 1; break; }case 7:{ xt = x - 1; yt = y + 2; break; }case 8:{ xt = x + 1; yt = y + 2; break; }}return;}stack<node>stk;bool dfs(int step, int x, int y){mark[x][y] = 1;node u;u.r = x;u.c = 'A' + y - 1;stk.push(u);if (step == p*q)return true;for (int i = 1; i <= 8; i++){move(i, x, y);if (isillegal(xt, yt)){if (dfs(step + 1, xt, yt))return true;}}stk.pop();mark[x][y] = 0;return false;}int main(){int casen;cin >> casen;for (int cas = 1; cas <= casen; cas++){cin >> p >> q;int ok = 0;memset(mark, 0, sizeof(mark));for (int i = 1; i <= p; i++){for (int j = 1; j <= q; j++){if (dfs(1, i, j)){ok = 1; break;}}if (ok) break;}if (cas != 1) cout << endl;cout << "Scenario #" << cas << ":" << endl;if (ok){int i = 0;node ans[900];while (!stk.empty()){ans[i++] = stk.top();stk.pop();}for (int j = i - 1; j >= 0; j--)cout << ans[j].c << ans[j].r;cout << endl;}else cout << "impossible" << endl;}}