Codeforces Round #165 (Div. 1) Greenhouse Effect（DP）

Greenhouse Effect

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

【题意】给你n朵花共m种，在实数坐标轴上依次排开。现在要移动最少的花的数量，使得重排后的花相同种类逇相邻，且从左到右种类编号是1~n.

【分析】DP。设dp[i][j]表示前i个排列好以编号为j的物种结尾所移动的最少次数，对于a[i+1]，如果a[i+1]>=j，那么a[i+1]可以不动也可以移到后面某一正确位置；如果a[i+1]<j则将a[i+1]移动到前面正确位置。转移方程为：

if  a[i]>=j
dp[i][a[i]] = min(dp[i - 1][j], dp[i][a[i]]); dp[i][j] = min(dp[i - 1][j]+1, dp[i][j]);
else  dp[i][j] = min(dp[i - 1][j]+1, dp[i][j]);

其实可以直接求最长非减子序列，这里麻烦了。

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
using namespace std;
typedef long long LL;
const int N = 5e3+;
const int mod = 1e9+;
int n,m;
int dp[N][N],a[N];
int main(){
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;i++){
        double x;
        scanf("%d%lf",&a[i],&x);
    }
    met(dp,inf);
    dp[][]=;
    for(int i=;i<=n;i++){
        for(int j=;j<=a[i];j++){
            dp[i][a[i]]=min(dp[i][a[i]],dp[i-][j]);
            dp[i][j]=min(dp[i][j],dp[i-][j]+);
        }
        for(int j=a[i]+;j<=m;j++){
            dp[i][j]=min(dp[i][j],dp[i-][j]+);
        }
    }
    int ans=inf;
    for(int i=;i<=m;i++){
        ans=min(ans,dp[n][i]);
    }
    printf("%d\n",ans);
    return ;
}