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：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <cmath>
6 #include <string>
7 #include <vector>
8 #include <set>
9 #include <map>
10 #include <queue>
11 #include <stack>
12 #include <sstream>
13 #include <cctype>
14 #include <cassert>
15 #include <typeinfo>
16 #include <utility> //std::move()
17 using std::cin;
18 using std::cout;
19 using std::endl;
20 const int INF = 0x7fffffff;
21 const double EXP = 1e-8;
22 const int MS = 100005;
23 typedef long long LL;
24 int left[MS], right[MS];
25 int n, m, kase;
26 void link(int l, int r)
27 {
28 right[l] = r;
29 left[r] = l;
30 }
31
32 int main()
33 {
34 int kase = 0;
35 while (cin >> n >> m)
36 {
37 for (int i = 1; i <= n; i++)
38 {
39 left[i] = i - 1;
40 right[i] = (i + 1) % (n + 1);
41 }
42 right[0] = 1;
43 left[0] = n;
44 int op, x, y, inv = 0;
45 while (m--)
46 {
47 cin >> op;
48 if (op == 4)
49 inv = !inv;
50 else
51 {
52 cin >> x >> y;
53 if (op == 3 && right[y] == x)
54 std::swap(x, y);
55 if (op != 3 && inv)
56 op = 3 - op;
57 if (op == 1 && x == left[y])
58 continue;
59 if (op == 2 && x == right[y])
60 continue;
61 int lx = left[x], rx = right[x], ly = left[y], ry = right[y];
62 if (op == 1)
63 {
64 link(lx, rx);
65 link(ly, x);
66 link(x, y);
67 }
68 else if (op == 2)
69 {
70 link(lx, rx);
71 link(y, x);
72 link(x, ry);
73 }
74 else if (op == 3)
75 {
76 if (right[x] == y)
77 {
78 link(lx, y);
79 link(y, x);
80 link(x, ry);
81 }
82 else
83 {
84 link(lx, y);
85 link(y, rx);
86 link(ly, x);
87 link(x, ry);
88 }
89 }
90 }
91
92 }
93 int b = 0;
94 LL ans = 0;
95 for (int i = 1; i <= n; i++)
96 {
97 b = right[b];
98 if (i % 2)
99 ans += b;
100 }
101 if (inv&&n % 2 == 0)
102 ans = (LL)n*(n + 1) / 2 - ans;
103 cout << "Case " << ++kase << ": " << ans << endl;
104 }
105 return 0;
106 }