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-优快云博客

本文链接：https://blog.youkuaiyun.com/ly59782/article/details/78077756

本文介绍了一种寻找序列中最长增重子序列的算法实现，通过对特定数值范围内的整数进行权重分配，利用动态规划思想高效求解最大权重和对应的非降子序列。

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Let $S$ be a sequence of integers $s_{1}$ , $s_{2}$ , $. . .$ , $s_{n}$ 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 $s_{i}$ is $s_{i}-10000$ . For example, if $s_{i}$ 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 $s_{i1}$ , $s_{i2}$ , $. . .$ , $s_{ik}$ , with $i_{1}<i_{2}\ ...\ <i_{k}$ , such that, for all $\leq j<k$ , we have $s_{ij}<s_{ij+1}$ .

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 $2*10^{5}$

Input Format

A list of integers separated by blanks: $s_{1}$ , $s_{2}$ , $. . .$ , $s_{n}$

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

/*
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;
}