A Knight’s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 58714 Accepted: 19998
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
Source
TUD Programming Contest 2005, Darmstadt, Germany
問題鏈接:POJ2488 A Knight’s Journey
問題簡述:給定國際棋盤的大小(p×q),判定馬能否不重複地跳過所有格,輸出其字典序順序的第一種路徑。
問題分析:經典的“騎士遊歷”問題,關鍵是跳馬的順序,保證先跳到字典順序的點。需要注意行按數字輸出列按字母輸出。
程序說明:(略)
參考鏈接:(略)
題記:(略)
AC的C++語言程序如下:
/* POJ2488 A Knight's Journey */
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int DN = 8;
// 字典順序方向
const int dx[] = {-1, 1, -2, 2, -2, 2, -1, 1};
const int dy[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int N = 26 + 1;
bool vis[N][N];
struct Setp {
char x, y;
} path[N];
int p, q, success;
void dfs(int x, int y, int step)
{
path[step].y = y + 'A' - 1; //int 轉爲 char
path[step].x = x + '0';
if (step == p * q) success = true;
else {
for (int i = 0; i < DN && !success; i++) {
int nx = x + dx[i];
int ny = y + dy[i];
if (0 < nx && nx <= p && 0 < ny && ny <= q &&
!vis[nx][ny] && !success) {
vis[nx][ny] = true; // 設置狀態,避免回跳
dfs(nx, ny, step + 1);
vis[nx][ny] = false; // 恢復狀態
}
}
}
}
int main()
{
int t;
scanf("%d", &t);
for(int k = 1; k <= t; k++) {
scanf("%d%d", &p, &q);
memset(vis, false, sizeof(vis));
vis[1][1] = true; // 起點
success = false;
dfs(1, 1, 1);
printf("Scenario #%d:\n", k);
if(success) {
for(int i = 1; i <= p * q; i++)
printf("%c%c", path[i].y, path[i].x);
printf("\n");
} else printf("impossible\n");
if(k != t) printf("\n"); // 空一行
}
return 0;
}