Codeforces Round #335 (Div. 2) C. Sorting Railway Cars 动态规划-优快云博客

本文介绍了一个火车车厢排序问题，目标是最少次数的操作使车厢编号升序排列。操作包括选取一个车厢并将其放置到列车的开头或末尾。文章提供了解决方案及代码实现。

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C. Sorting Railway Cars

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://www.codeforces.com/contest/606/problem/C

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 pi (1 ≤ pi ≤ n, pi ≠ pj 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

题意

给你1-n的一个排列

然后你的操作是可以把一个数放到结尾或者首部

然后问你最少多少次操作，可以将这个排列变成递增的

题解：

首先跑最长上升子序列是错的

比如 1 2 4 5 3

最长上升子序列答案跑出来是1

但实际上答案是2

所以只能跑严格只比之前大1的子序列就好了

代码：

#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cstdio>
using namespace std;

int dp[100005];
int a[100005];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        dp[a[i]]=dp[a[i]-1]+1;
    int flag = 0;
    for(int i=1;i<=n;i++)
        flag = max(flag,dp[i]);
    cout<<n-flag<<endl;
}