移动盒子（Boxes in a Line）

You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4kinds of commands:
• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )
• 2 X Y : move box X to the right to Y (ignore this if X is already the right of Y )
• 3 X Y : swap box X and Y
• 4: reverse the whole line.
Commands are guaranteed to be valid, i.e. X will be not equal to Y .For example, if n = 6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1.Then after executing 4, then line becomes 1 3 5 4 6 2

Input
There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m(1 ≤ n, m ≤ 100, 000). Each of the following m lines contain a command.

Output
For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to nfrom left to right.

Sample Input
6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4

Sample Output
Case 1: 12
Case 2: 9
Case 3: 2500050000

分析：

使用双向链表，用left_[i]和right_[i]分别表示编号位i的盒子左边和右边的盒子编号（如果是0，表示不存在），通过以下过程可以将两个结点相互连接：

void link(int a, int b)
{
    
    
	right_[a] = b;
	left_[b] = a;
}

注意：
操作4比较特殊，为了避免一次修改所有元素的指针，此处添加一个标记flat，表示有没有执行过操作4，（若flare = 1时，再执行一次操作4，则flat变为0）。这样，当op为1或2且flat = 1时，只需吧op变成3-op（注意操作3不受flat的影响）即可。最终输出时根据flat的值的不同进行不同的处理。

源代码：

#include<iostream>
using namespace std;
const int maxn = 100000 + 5;
int left_[maxn], right_[maxn];
void link(int a, int b);
int main()
{
    
    
	int n, m, k = 1;
	int op, flat = 0;
	while (scanf("%d%d", &n, &m) == 2)
	{
    
    
		flat = 0;
		for (int i = 1; i <= n; i++)
		{
    
    
			left_[i] = i - 1;
			right_[i] = i + 1;
		}
		right_[n] = 0;
		right_[0] = 1;
		left_[0] = n;
		while(m --)
		{
    
    
			scanf("%d", &op);
			if (op == 4) flat = !flat;
			else
			{
    
    
				int X, Y;
				scanf("%d%d", &X, &Y);
				if (op != 3 && flat) op = 3 - op;
				if (op == 1 && left_[Y] == X) continue;
				if (op == 2 && right_[Y] == X) continue;

				int LX = left_[X], RX = right_[X], LY = left_[Y], RY = right_[Y];
				if (op == 1)
				{
    
    
					link(LX, RX);
					link(X, Y);
					link(LY, X);
				}
				else if (op == 2)
				{
    
    
					link(LX, RX);
					link(Y, X);
					link(X, RY);
				}
				else if (op == 3)
				{
    
    
					if (right_[Y] == X)
						X ^= Y ^= X ^= Y;
					if (right_[X] == Y)
					{
    
    
						link(LX, Y); link(Y, X); link(X, RY);
					}
					else
					{
    
    
						link(LX, Y); link(Y, RX); link(LY, X); link(X, RY);
					}
				}
			}
		}
		int b = 0;
		long long ans = 0;
		for (int i = 1; i <= n; i++)
		{
    
    
			b = right_[b];
			if(i % 2 == 1)
				ans += b;
		}
		if (n % 2 == 0 && flat == 1)
		{
    
    
			ans = n * (n + 1) / 2 - ans;
		}
		cout << "Case " << k++ << ": " << ans << endl;
	}
	return 0;
}
void link(int a, int b)
{
    
    
	right_[a] = b;
	left_[b] = a;
}