Boxes in a Line

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 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

直接模拟肯定会超时 用stl中的链表也超时 只能用数组自己模拟一个双向链表了 le[i],ri[i]分别表示第i个盒子左边盒子的序号和右边盒子的序号

参考代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <sstream>
#include <cctype>
#include <cassert>
#include <typeinfo>
#include <utility>    //std::move()
using std::cin;
using std::cout;
using std::endl;
const int INF = 0x7fffffff;
const double EXP = 1e-8;
const int MS = 100005;
typedef long long LL;
int left[MS], right[MS];
int n, m, kase;
void link(int l, int r)
{
    right[l] = r;
    left[r] = l;
}

int main()
{
    int kase = 0;
    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)
                    std::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;
        LL ans = 0;
        for (int i = 1; i <= n; i++)
        {
            b = right[b];
            if (i % 2)
                ans += b;
        }
        if (inv&&n % 2 == 0)
            ans = (LL)n*(n + 1) / 2 - ans;
        cout << "Case " << ++kase << ": " << ans << endl;
    }
    return 0;
}