[Codeforces 606C]Sorting Railway Cars

最新推荐文章于 2018-07-12 11:40:42 发布

转载最新推荐文章于 2018-07-12 11:40:42 发布 · 127 阅读

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原文链接：http://www.cnblogs.com/NaVi-Awson/p/7681131.html

本文介绍了一个火车排序问题：如何通过将火车车厢传送到首尾来最小化操作次数以完成排序。通过对问题的深入分析，提出了一种高效算法，能够在O(n)的时间复杂度内解决问题。文章还提供了一份简洁清晰的C++实现代码。

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Description

An infinitely long railway has a train consisting of n cars, numbered from 1 to n (the numbers of all the cars are distinct) and positioned in arbitrary order. David Blaine wants to sort the railway cars in the order of increasing numbers. In one move he can make one of the cars disappear from its place and teleport it either to the beginning of the train, or to the end of the train, at his desire. What is the minimum number of actions David Blaine needs to perform in order to sort the train?

Input

The first line of the input contains integer n (1 ≤ n ≤ 100 000) — the number of cars in the train.

The second line contains n integers p_i (1 ≤ p_i ≤ n, p_i ≠ p_j if i ≠ j) — the sequence of the numbers of the cars in the train.

Output

Print a single integer — the minimum number of actions needed to sort the railway cars.

Sample Input

5
4 1 2 5 3

Sample Output

Hint

In the first sample you need first to teleport the 4-th car, and then the 5-th car to the end of the train.

题解

首先容易注意到答案一定不会超过 $n$,因为我们只要按照大小顺序每个数都移动一次,整个序列就一定有序了,由以上结论还可以得每个数至多被移动一次。

由于操作是将数移动到序列首部和序列尾部,那么没有被移动的数会停留在序列中部,所以这些数一定是原序列的一个子序列,并且是有序序列的一个连续子段。最小化移动即是最大化不动,那么我们找到最长的这样的一个子序列就行了。

时间复杂度 $O(n)$ 。

 1 //It is made by Awson on 2017.10.17
 2 #include <set>
 3 #include <map>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <queue>
 7 #include <stack>
 8 #include <vector>
 9 #include <cstdio>
10 #include <string>
11 #include <cstdlib>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 #define LL long long
16 #define Max(a, b) ((a) > (b) ? (a) : (b))
17 #define Min(a, b) ((a) < (b) ? (a) : (b))
18 using namespace std;
19 const int N = 200000;
20 void read(int &x) {
21   char ch = getchar(); x = 0;
22   while (ch < '0' || ch > '9') ch = getchar();
23   while (ch >= '0' && ch <= '9') x = (x<<1)+(x<<3)+ch-48, ch = getchar();
24 }
25 
26 int n, ans, a;
27 int f[N+5];
28 
29 void work() {
30   read(n);
31   for (int i = 1; i <= n; i++) {
32     read(a); f[a] = f[a-1]+1;
33     ans = Max(ans, f[a]);
34   }
35   printf("%d\n", n-ans);
36 }
37 int main() {
38   work();
39   return 0;
40 }

转载于:https://www.cnblogs.com/NaVi-Awson/p/7681131.html