UVA 12657 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
1 /*
2 Uva 12657
3 双向链表
4 */
5 #include<iostream>
6 #include<cstdio>
7 #include<algorithm>//swap函数在其内部
8 using namespace std;
9 const int maxn = 100000 + 5;
10 int r[maxn], l[maxn];
11 void link(int x, int y)
12 {
13 r[x] = y, l[y] = x;
14 }
15 int main()
16 {
17 int n, signify, kase = 0;
18 while (scanf("%d%d", &n, &signify) == 2)
19 {
20 for (int i = 1; i <= n; i++)
21 {
22 l[i] = i - 1;
23 r[i] = (i + 1) % (n + 1);
24 }
25 r[0] = 1;
26 l[0] = n;
27 int op, x, y, mark = 0;
28 while (signify--)
29 {
30 scanf("%d", &op);
31 if (op == 4)mark = !mark;
32 else
33 {
34 scanf("%d%d", &x, &y );
35 if (op == 3 && r[y] == x)swap(x, y);//确保x在y的左边,统一一下，后面方便处理
36 if (op != 3 && mark)op = 3 - op;
37 if (op == 1 && x == l[y])continue;
38 if (op == 2 && x == r[y])continue;
39 int lx = l[x], rx = r[x], ly = l[y], ry = r[y];//在连接节点的过程中，原始值会发生改变l[x],r[x]之类的会因为link函数发生改变，会导致之后系列的连接出错，所以要存储其初始值
40 if (op == 1)
41 {
42 link(lx, rx);//x的上一个节点和下一个节点形成双向链表
43 link(ly, x);//y的上一个节点和x形成双向链表
44 link(x, y);//y的上一个节点和y形成双向链表
45 }
46 else if (op == 2)
47 {
48 link(lx, rx);
49 link(x, ry);
50 link(y, x);
51 }
52 else if (op == 3)
53 {
54 if (r[x] == y) { link(lx, y); link(y, x); link(x, ry); }
55 else { link(lx, y); link(y, rx); link(ly, x); link(x, ry); }
56 }
57 }
58 }
59 int b = 0;
60 long long ans = 0;
61 for (int i = 1; i <= n; i++)
62 {
63 b = r[b];
64 if (i % 2 == 1)ans += b;
65 }
66 if (mark && n % 2 == 0)ans = (long long)(n * ((long long)n + 1)) / 2 - ans;//考虑n的奇偶数
67 printf("Case %d: %lld\n", ++kase, ans);
68 }
69
70 return 0;
71 }
原文地址：https://www.cnblogs.com/Dinosaur-Po/p/12685444.html