codeforces269B

最新推荐文章于 2020-04-02 21:40:46 发布

转载最新推荐文章于 2020-04-02 21:40:46 发布 · 168 阅读

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原文链接：http://www.cnblogs.com/gaojunonly1/p/10798587.html

探讨在无限长的温室中，如何通过最少的植物移栽，将不同物种的植物按顺序分区种植的问题。使用最长不下降子序列算法解决挑战。

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Greenhouse Effect

CodeForces - 269B

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 s_i (1 ≤ s_i ≤ m), and one real number x_i (0 ≤ x_i ≤ 10⁹), the species and position of the i-th plant. Each x_i will contain no more than 6 digits after the decimal point.

It is guaranteed that all x_i 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 x_i coordinates (x_i < x_i + 1, 1 ≤ i < n).

Output

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

Examples

Input

Output

Input

Output

Input

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

Output

2

sol:题意略坑，按照颜色跑一遍最长不下降子序列就好了

//给出一个实数轴，上面散布着n个点，他们是m种，问最少挪多少步能将数轴上的点分成1~m的种类顺序排列的m块。
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=5005;
int n,m,dp[N];
struct Point
{
    int Cor;
    double Pos;
}P[N];
int main()
{
    int i,j,ans=0;
    R(n); R(m);
    for(i=1;i<=n;i++)
    {
        R(P[i].Cor); scanf("%lf",&P[i].Pos);
    }
    P[0].Cor=0; dp[0]=0;
    for(i=1;i<=n;i++)
    {
        for(j=0;j<i;j++) if(P[j].Cor<=P[i].Cor) dp[i]=max(dp[i],dp[j]+1);
        ans=max(ans,dp[i]);
    }
    Wl(n-ans);
    return 0;
}
/*
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

Input
15 5
4 6.039627
2 7.255149
2 14.469785
2 15.108572
4 22.570081
5 26.642253
5 32.129202
5 44.288220
5 53.231909
5 60.548042
4 62.386581
2 77.828816
1 87.998512
3 96.163559
2 99.412872
Output
6
*/

View Code

转载于:https://www.cnblogs.com/gaojunonly1/p/10798587.html