Uva12657 (Boxes in a Line)移動盒子

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     雙向鏈表
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<algorithm>//swap函數在其內部
 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);//確保x在y的左邊,統一一下，後面方便處理
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];//在鏈接節點的過程當中，原始值會發生改變l[x],r[x]之類的會由於link函數發生改變，會致使以後系列的鏈接出錯，因此要存儲其初始值
40                 if (op == 1)
41                 {
42                     link(lx, rx);//x的上一個節點和下一個節點造成雙向鏈表
43                     link(ly, x);//y的上一個節點和x造成雙向鏈表
44                     link(x, y);//y的上一個節點和y造成雙向鏈表
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;//考慮n的奇偶數
67         printf("Case %d: %lld\n", ++kase, ans);
68     }
69 
70     return 0;
71 }