C++——【USACO 4.4.1】——Shuttle Puzzle

Shuttle Puzzle
Traditional

The Shuttle Puzzle of size 3 consists of 3 white marbles, 3 black marbles, and a strip of wood with 7 holes. The marbles of the same color are placed in the holes at the opposite ends of the strip,
leaving the center hole empty.

INITIAL STATE: WWW_BBB
GOAL STATE: BBB_WWW

To solve the shuttle puzzle, use only two types of moves. Move 1 marble 1 space (into the empty hole) or jump 1 marble over 1 marble of the opposite color (into the empty hole). You may not back up,
and you may not jump over 2 marbles.
A Shuttle Puzzle of size N consists of N white marbles and N black marbles and 2N+1 holes.
Here's one solution for the problem of size 3 showing the initial, intermediate, and end states:

WWW BBB
WW WBBB
WWBW BB
WWBWB B
WWB BWB
W BWBWB
WBWBWB
BW WBWB
BWBW WB
BWBWBW
BWBWB W
BWB BWW
B BWBWW
BB WBWW
BBBW WW
BBB WWW

Write a program that will solve the SHUTTLE PUZZLE for any size N (1 <= N <= 12) in the minimum number of moves and display the successive moves, 20 per line.

PROGRAM NAME: shuttle

INPUT FORMAT

A single line with the integer N.

SAMPLE INPUT (file shuttle.in)

3

OUTPUT FORMAT

The list of moves expressed as space-separated integers, 20 per line (except possibly the last line). Number the marbles/holes from the left, starting with one.
Output the solution that would appear first among the set of minimal solutions sorted numerically (first by the first number, using the second number for ties, and so on).

SAMPLE OUTPUT (file shuttle.out)

3 5 6 4 2 1 3 5 7 6 4 2 3 5 4

有道翻译得太奇怪，这里的翻译从其他博客上copy的。

描述

大小为3的棋盘游戏里有3个白色棋子，3个黑色棋子，和一个有7个格子一线排开的木盒子。3个白棋子被放在一头，3个黑棋子被放在另一头，中间的格子空着。

初始状态: WWW_BBB
目标状态: BBB_WWW

在这个游戏里有两种移动方法是允许的：
你可以把一个棋子移到与它相邻的空格；
你可以把一个棋子跳过一个(仅一个)与它不同色的棋子到达空格。

大小为N的棋盘游戏包括N个白棋子，N个黑棋子，还有有2N+1个格子的木盒子。

这里是3-棋盘游戏的解，包括初始状态，中间状态和目标状态：

 WWW BBB
WW WBBB
WWBW BB
WWBWB B
WWB BWB
W BWBWB
WBWBWB
BW WBWB
BWBW WB
BWBWBW
BWBWB W
BWB BWW
B BWBWW
BB WBWW
BBBW WW
BBB WWW

请编一个程序解大小为N的棋盘游戏(1 <= N <= 12)。要求用最少的移动步数实现。

格式

PROGRAM NAME: shuttle

INPUT FORMAT:

(file shuttle.in)

一个整数N。

OUTPUT FORMAT:

(file shuttle.out)

输出用移动的棋子在棋盘的位置(位置从左到右依次为1, 2, ..., 2N+1)表示的变换序列，每
4000
个数字之间以空格分隔，每行20个数(除了最后一行)。

输出的解还应当有最小的字典顺序(即如果有多组移动步数最小的解，输出第一个数最小的解；如果还有多组，输出第二个数最小的解；...)。

SAMPLE INPUT

3

SAMPLE OUTPUT

3 5 6 4 2 1 3 5 7 6 4 2 3 5 4

第一眼以为是暴力dfs, 然而........向那位找规律的大佬低头...

ps : 被三目坑了....那玩意儿简直有毒......

/*
ID : mcdonne1
LANG : C++
TASK : shuttle
*/

#pragma GCC optimize("O3")

#include <cstdio>

using namespace std;

int a[13][13];

int main () {
freopen ("shuttle.in", "r", stdin);
freopen ("shuttle.out", "w", stdout);
int n, cnt = 0, start = 0, it;
scanf ("%d", &n);
a[1][1] = n + 1;
a[1][0] = 1;
for (int i = 2; i <= n + 1; i++) {
a[i][0] = a[i - 1][0] + 1;
if (a[i - 1][1] & 1) {
if (n % 2) it = -2, start = n - i + 2 + (a[i][0] << 1);
else it = 2, start = n - i;
for (int j = 1; j <= a[i][0]; j++) {
printf ("%d", a[i][j] = start += it);
putchar ((++cnt) % 20 ? ' ' : '\n');
}
}
else {
if (n % 2) it = 2, start = n + i - 2 * a[i][0];
else it = -2, start = n + i + 2;
for (int j = 1; j <= a[i][0]; j++) {
printf ("%d", a[i][j] = start += it);
putchar ((++cnt) % 20 ? ' ' : '\n');
}
}
}
for (int i = n; i > 1; i--)
for (int j = 1; j <= a[i][0]; j++) {
printf ("%d", a[i][j]);
putchar ((++cnt) % 20 ? ' ' : '\n');
}
printf ("%d\n", n + 1);
return 0;
}