【Codeforces 605A】【貪心】Sorting Railway Cars

Sorting Railway Cars

Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

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.

Samples

Input1

5
4 1 2 5 3

Output1

2

Input2

4
4 1 3 2

Output2

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.

Source

Codeforces Round #335 (Div. 1)

就是一道貪心的題，題目中說了序號，所以序號一定是<=n的，不用排序然後怎麼再來搞，直接O(n)就可以求LIS了

代碼見下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<stack>
#define INF 2100000000
#define ll long long
#define clr(x)  memset(x,0,sizeof(x))
#define clrmax(x)  memset(x,127,sizeof(x))

using namespace std;

inline int read()
{
    char c;
    int ret=0;
    while(!(c>='0'&&c<='9'))
        c=getchar();
    while(c>='0'&&c<='9')
    {
        ret=(c-'0')+(ret<<1)+(ret<<3);
        c=getchar();
    }
    return ret;
}

#define M 100005
int n,f[M],a[M];

int main()
{
    n=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    for(int i=1;i<=n;i++)
        f[a[i]]=f[a[i]-1]+1;
    int mx=0;
    for(int i=1;i<=n;i++)mx=max(mx,f[i]);
    cout<<n-mx;
    return 0;

}

大概就是這個樣子，如果有什麼問題，或錯誤，請在評論區提出，謝謝。