Boxes in a Line UVA - 12657

  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

HINT

  这个题重点就是如何解决超时的问题，如果老老实实的按照题目的要求来反转链表，那超时是肯定的了。另外利用函数查找也是超时。一开始值考虑到反转超时，没考虑查找超时问题，看了vj上一个大佬的思路用数组储存地址。因为链表只有反转后和反转前两种状态，而当链表为反转时，x放到y的右侧就是真的右侧，而反转后便是y的左侧。因此只需要一个标志位记录链表的状态就好。等到最后输出的时候在看链表时反转的还是正向的，进行选择反转。然后计算答案输出。

Accepted

#include<bits/stdc++.h>
using namespace std;

int main() {
	int m, n, op, x, y, num = 0;
	while (cin >> m >> n) {
		int q = 1;
		list<int>L(m);			//这里必须先说明大小为m否则如果m过大会卡死程序，不知道为啥~
		vector<list<int>::iterator>pointer(m+1);	//这里必须先说明大小为m否则如果m过大会卡死程序，不知道为啥~
		for (auto temp = L.begin();temp != L.end();temp++) {	//初始化
			*temp = q++;
			pointer[q - 2] = temp;
		}
		int flag = 1;		//判断正反1为正，-1为反
		while (n--) {
			cin >> op;
			if (op == 4)flag = -1 * flag;		//标记反转
			else {
				cin >> x >> y;
				x--;y--;
				if (op == 3) {					//交换的时候要值和地址一起交换
					swap(*pointer[x], *pointer[y]);
					swap(pointer[x], pointer[y]);
				}
				else {
					auto temp1 = pointer[x];		//先插到y的右面
					pointer[x] = L.insert(pointer[y], *pointer[x]);
					L.erase(temp1);
					if (flag == 1 && op == 2 || flag == -1 && op == 1) {	//如果要查到右边就交换x,y
						swap(*pointer[x], *pointer[y]);
						swap(pointer[x], pointer[y]);
					}
				}
			}
		}
			unsigned long long int sum = 0;			//计算
			auto temp = L.begin();
			if (flag == -1)temp = --L.end();
			m = L.size() - 1;
			for (int i = 0;i < L.size();i++, m--)
			{
				if (i % 2 == 0)sum += *temp;
				if (flag == 1)temp++;				//对反转还是正向进行区分
				else if (temp != L.begin()) temp--;
			}
			cout << "Case " << ++num << ": " << sum << endl;
		

	}
}