L. The Heaviest Non-decreasing Subsequence Problem -最长不降子序列变形nlogn-2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

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.

样例输入

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

样例输出

14

/*题意：给你一个数列，当该数数值<0时该数无价值，当该数数值>=10000时 该数值-=10000，其价值为5，其余情况价值为1问价值和最大的非降子序列是多少题解：价值可以转换为长度>=10000的数，就在数组中存5次<0的数因为没价值，并且求得是子序列不是子串，不影响最后序列的选取，所以根本不用存最后求得的长度即为答案*/#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;}