非连续递增序列

最新推荐文章于 2025-02-27 22:20:00 发布

Qingzhx

最新推荐文章于 2025-02-27 22:20:00 发布

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本文链接：https://blog.youkuaiyun.com/Qingzhx/article/details/115642020

该博客介绍了一种解决CodeForces-606C问题的方法，即如何用最少的操作次数将一列火车车厢按编号从小到大排序。通过建立辅助数组并计算每个车厢位置的偏移量，可以确定所需操作次数。在给出的两个示例中，都只需要进行两次操作即可完成排序。此问题涉及贪心算法和数组处理技巧。

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CodeForces - 606C
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.

Examples
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.

在数据较小时，利用一个数组，以数值为下标来解答

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[100001]={0}, b[100001]={0};
int main()
{
    int n, i,maxn=0;
    cin >> n;
    for (i = 1; i <= n; i++)
        cin >> a[i];
    for (i = 1; i <= n; i++)
    {
        b[a[i]] = b[a[i] - 1] + 1;//会根据输入先后变化取值，
        //正确地得出序列长度
        maxn = max(maxn, b[a[i]]);
    }
    cout << n - maxn << endl;
    system("pause");
    return 0;
}