You have n boxes in a line on the table numbered 1...n from left to right. Your task is to simulate 4 kinds 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 executing 2 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 n from 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

思路：采用双向链表：用left[i]和right[i]分别表示编号为i的盒子左边和右边的盒子编号（如果是0，表示不存在），再使用link函数将两个节点连接

void link(int L, int R)
{
	Right[L] = R; Left[R] = L;
}

但操作4需另外讨论，为避免一次修改所有元素的指针，可增加一标记来表示是否执行过4.

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#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;
#define MAXN 100010
int Left[MAXN], Right[MAXN];
int m, n, cas;

void link(int L, int R)
{
	Right[L] = R; Left[R] = L;
}
int main()
{
	while (cin >> n >> m)
	{
		for (int i = 1; i <= n; i++)
		{
			Left[i] = i - 1;
			Right[i] = (i + 1) % (n + 1);
		}
		Right[0] = 1; Left[0] = n;

		int op, x, y, inv = 0;
		while (m--)
		{
			cin >> op;
			if (op == 4)
				inv = !inv;
			else
			{
				cin >> x >> y;
				if (op == 3 && Right[y] == x)swap(x, y);
				if (op != 3 && inv)op = 3 - op;
				if (op == 1 && x == Left[y])continue;
				if (op == 2 && x == Right[y])continue;

				int lx = Left[x], rx = Right[x], ly = Left[y], ry = Right[y];
				if (op == 1)
				{
					link(lx, rx); link(ly, x); link(x, y);
				}
				else if (op == 2)
				{
					link(lx, rx); link(y, x); link(x, ry);
				}
				else if (op == 3)
				{
					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 (inv&&n % 2 == 0)
			ans = (long long)n*(n + 1) / 2 - ans;
		printf("Case %d: %lld\n", ++cas, ans);
	}
	return 0;
}