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：
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 }
