Codeforces270D Greenhouse Effect【思维+LIS】

最新推荐文章于 2021-03-16 20:23:40 发布

mengxiang000000

最新推荐文章于 2021-03-16 20:23:40 发布

阅读量532

点赞数

CC 4.0 BY-SA版权

分类专栏： dp 思维文章标签： Codeforces270D

本文链接：https://blog.youkuaiyun.com/mengxiang000000/article/details/78164965

思维同时被 2 个专栏收录

692 篇文章

订阅专栏

351 篇文章

订阅专栏

解决一个关于植物重新布局的问题，通过移动最少数量的植物来确保不同种类的植物按升序排列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.

Over the years Emuskald has cultivated n plants in his greenhouse, of m different plant species numbered from 1 to m. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.

Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange m - 1 borders that would divide the greenhouse into m sections numbered from 1 to m from left to right with each section housing a single species. He is free to place the borders, but in the end all of the i-th species plants must reside in i-th section from the left.

Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.

Input

The first line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 5000, n ≥ m), the number of plants and the number of different species. Each of the following n lines contain two space-separated numbers: one integer number si (1 ≤ si ≤ m), and one real number xi (0 ≤ xi ≤ 109), the species and position of the i-th plant. Each xi will contain no more than 6 digits after the decimal point.

It is guaranteed that all xi are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their xi coordinates (xi < xi + 1, 1 ≤ i < n).

Output

Output a single integer — the minimum number of plants to be replanted.

Examples

input
3 2
2 1
1 2.0
1 3.100

output
1

input
3 3
1 5.0
2 5.5
3 6.0

output
0

input
6 3
1 14.284235
2 17.921382
1 20.328172
3 20.842331
1 25.790145
1 27.204125

output
2

Note

In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.

In the second test case, the species are already in the correct order, so no replanting is needed.

题目大意：

给出N个植物，每个植物都属于一个品种，共计m个品种，分落在不同的位子上（在一个数轴上，而且数轴是无限长度的），保证读入的位子是按照升序读入的。

现在我们可以进行一个操作：取任意一个位子上的植物，移动到任意一个没有植物的位子上去。

问我们最少进行多少次操作，能够使得从左到右，是按照品种升序排列的（1~m），而且每种植物都相邻；

思路：

①仔细读题之后发现，位子是没有用的，那么我们的任务就是挪动最少次数，使得序列品种升序。

②再仔细思考一下，我们挪动到的位子是不限的，那么我们其实问题就是确定最多多少个植物不挪动相对位子，而挪动其他植物即可，

那么问题就是求一个LIS，答案就是n-Lis

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int a[150000];
int have[150000];
int ned[150000];
int dp[150000];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(have,0,sizeof(have));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            double pos;
            scanf("%d%lf",&a[i],&pos);
        }
        int maxn=0;
        for(int i=1;i<=n;i++)
        {
            dp[i]=1;
            for(int j=i-1;j>=1;j--)
            {
                if(a[i]>=a[j])
                {
                    dp[i]=max(dp[i],dp[j]+1);
                }
            }
            maxn=max(maxn,dp[i]);
        }
        printf("%d\n",n-maxn);
    }
}