POJ 2488 A Knight's Journey__深搜

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

分析

题目大意：骑士要在国际象棋上走，见描述中的图，骑马只能走日字，问有无办法每个点只走一次走完全部棋盘。只用搜索一种可行方案呗，那就从第一个点出发向8个方向搜索，棋盘坐标横向为字母A开始，纵向为数字1开关，需要注意的是搜索的方法要按字典序来，横向字母从小到大加纵向数字从小到大。

实现

#include <iostream>
#include <cstring>
#define PATH_NUM 51
using namespace std;
int visited[26][26];
char path[PATH_NUM][2];
//方向只能是这个顺序，具体看*分析*节描述。
int to[8][2] = { { -2, -1}, {-2, 1}, {-1, -2}, {-1, 2}, {1, -2}, {1, 2}, {2, -1}, {2, 1} };
int cases, m, n, maxStep;
bool success = false;

//从(r, c)点开始再次跳跃，搜索下一个落脚点，当前走了step步。
void dfs(int r, int c, int step)
{
if (success) return;
path[step][0] = r + '1';
path[step][1] = c + 'A';
if (step == maxStep) {
success = true;
return;
}
for (int i = 0; i < 8; i++) {
int newR = r + to[i][1];
int newC = c + to[i][0];
if ((newC >= 0 && newC < n) && (newR >= 0 && newR < m) && !visited[newR][newC]) {
visited[newR][newC] = 1;
dfs(newR, newC, step+1);
visited[newR][newC] = 0;
}
}
}

int main()
{
//  freopen("in.txt", "r", stdin);
cin >> cases;
int index = 0;
while(cases--) {
cin >> m >> n;
maxStep = m * n;
memset(visited, 0, sizeof(visited));
memset(path, 0, sizeof(path));
success = false;
index++;
visited[0][0] = 1;
dfs(0, 0, 1);
cout << "Scenario #" << index << ":" << endl;
if (success) {
for(int i = 1; i <= maxStep; i++) {
cout << path[i][1] << path[i][0];
}
}else {
cout << "impossible";
}
cout << endl << endl;
}
return 0;
}