Codeforces Round #335 (Div. 2)-Sorting Railway Cars(求连续的上升序列的最大值)

C. Sorting Railway Cars

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 test(s)

input

5
4 1 2 5 3

output

2

input

4
4 1 3 2

output

2

Note

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.

思路：

  这题很容易误导别人，就是一题求普通最长上升子序列的题目，其实这是一道变形题，规定了要是连续的上升序列才行。所以只需要在自身的条件加一，再加上在自己后一位的数的值就行了，就一模拟题。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cmath>
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
typedef __int64 ll;
#define T 100005
#define mod 1000000007
int v[T];
int main()
{
#ifdef zsc
freopen("input.txt","r",stdin);
#endif
int n,i,j,k;
while(~scanf("%d",&n))
{
memset(v,0,sizeof(v));
for(i=0,j=1;i<n;++i){
scanf("%d",&k);
v[k] = v[k-1]+1;
j = max(j,v[k]);
}
printf("%d\n",n-j);
}
return 0;
}