L. The Heaviest Non-decreasing Subsequence Problem -最長不降子序列變形nlogn-2017 ACM-ICPC 亞洲區（南寧賽區）網絡賽

Let $S$ be a sequence of integers s1s_{1}s​1​​,s2s_{2}s​2​​,$...$,sns_{n}s​n​​ Each integer is is associated with a weight by the following rules:

(1) If is is negative, then its weight is $0$.

(2) If is is greater than or equal to $10000$, then its weight is $5$. Furthermore, the real integer value of sis_{i}s​i​​ is si−10000s_{i}-10000s​i​​−10000 . For example, if sis_{i}s​i​​ is $10101$, then is is reset to $101$ and its weight is $5$.

(3) Otherwise, its weight is $1$.

A non-decreasing subsequence of $S$ is a subsequence si1s_{i1}s​i1​​,si2s_{i2}s​i2​​,$...$,siks_{ik}s​ik​​, with i1<i2 ... <iki_{1}<i_{2}\ ...\ <i_{k}i​1​​<i​2​​ ... <i​k​​, such that, for all $1\le j<k$, we have sij<sij+1s_{ij}<s_{ij+1}s​ij​​<s​ij+1​​.

A heaviest non-decreasing subsequence of $S$ is a non-decreasing subsequence with the maximum sum of weights.

Write a program that reads a sequence of integers, and outputs the weight of its

heaviest non-decreasing subsequence. For example, given the following sequence:

$80$$75$$73$$93$$73$$73$$10101$$97$$-1$$-1$$114$$-1$$10113$$118$

The heaviest non-decreasing subsequence of the sequence is $<73,73,73,101,113,118>$ with the total weight being $1+1+1+5+5+1=14$. Therefore, your program should output $14$ in this example.

We guarantee that the length of the sequence does not exceed 2∗1052*10^{5}2∗10​5​​

Input Format

A list of integers separated by blanks:s1s_{1}s​1​​,s2s_{2}s​2​​,$...$,sns_{n}s​n​​

Output Format

A positive integer that is the weight of the heaviest non-decreasing subsequence.

樣例輸入

80 75 73 93 73 73 10101 97 -1 -1 114 -1 10113 118

樣例輸出

14

/*
題意：
給你一個數列，當該數數值<0時該數無價值，
當該數數值>=10000時 該數值-=10000，其價值爲5，
其餘情況價值爲1
問價值和最大的非降子序列是多少

題解：
價值可以轉換爲長度
>=10000的數，就在數組中存5次
<0的數因爲沒價值，並且求得是子序列不是子串，不影響最後序列的選取，所以根本不用存
最後求得的長度即爲答案
*/

#include <bits/stdc++.h>
#include <cstdio>

using namespace std;

int a[1000010];
int f[1000010];
int d[1000010];

int _bsearch(const int *f, int len, const int &a)
{
    int l = 0, r = len - 1;
    while(l <= r)
    {
        int mid = (l + r) / 2;
        if(a >= f[mid - 1]  && a < f[mid])
            return mid;
        else if(a < f[mid])
            r = mid - 1;
        else
            l = mid + 1;
    }
}

int LIS(const int *a, const int &n)
{
    int i,j,len = 1;
    f[0] = a[0];
    d[0] = 1;
    for(i = 1; i < n; i++)
    {
        if(a[i] < f[0])
            j = 0;
        else if(a[i] >= f[len - 1])
            j = len++;
        else
            j = _bsearch(f, len, a[i]);
        f[j] = a[i];
        d[i] = j + 1;
    }

    return len;
}

int main()
{
    int temp;
    int len = 0;

    while(cin>>temp)
    {
        if(temp > 0 && temp < 10000)
        {
            a[len++] = temp;
        }
        else if(temp >= 10000)
        {
            for(int i = 0; i < 5; i++)
                a[len++] = temp - 10000;
        }
    }
    int ans = LIS(a,len);
    cout<<ans<<endl;

    return 0;
}