贪心--Sorting Railway Cars

E - Sorting Railway Cars
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
SubmitStatus

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

Input

5
4 1 2 5 3

Output

2

Input

4
4 1 3 2

Output

2

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.

题意：对已经编号任意排列的车子进行排序，使它们按照编号从小到大排序，序列中的车子只能往两端移动；求最小移动的步数

思路：求最长连续子序列。

样例1：

4 1 2 5 3
最长连续子序列就是：* 1 2 * 3

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int dp[100005];
int a[100005];
int main()
{
//freopen("input.txt","r",stdin);
int n;
while(~scanf("%d",&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]);
printf("%d\n",n-flag);
}
return 0;
}