The Heaviest Non-decreasing Subsequence Problem（LIS+思维）

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本文介绍了一道算法题目，旨在寻找一个整数序列中非递减子序列的最大权值和，通过调整序列中元素的权值并利用最长非递减子序列的概念来解决。

题目链接：

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：给出一个序列，并赋予每个数权值，负数权值为0，0~9999权值为1，10000以上权值为5。求非减子序列的最大权值和，但是在判定是否为非减子序列时，最10000以上的值要减去10000，但其权值保持不变。

题解：负数权值为0，不用管。剩下的分为两个等级1和5，如果只有1或5，即每个数权值相同，那么只需求最长非减子序列即可。可是现在有两个值，怎么办呢？将5变成5个1！即把权值为5的数原地复制成5个相同的数，由于是非减序列，这样做并不会影响原来的序列，再求一个最长非减子序列即可。

#include <bits/stdc++.h>
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
int a[maxn];
int dp[maxn];

void Add(int k, int& i)     //在数组中加入这个值
{
    if(k >= 10000){
        k -= 10000;
        for(int j = 0; j < 5; j++) a[i++] = k;
    }
    else if(k > 0 && k < 10000) a[i++] = k;
}

int main()
{
    int k, i = 0;
    scanf("%d", &k);
    Add(k, i);
    while(getchar() == ' '){
        scanf("%d", &k);
        Add(k, i);
    }
    int n = i;

    fill(dp, dp + n, INF);
    for(int i = 0; i < n; i++){
        *upper_bound(dp, dp + n, a[i]) = a[i];
    }
    cout<<lower_bound(dp, dp + n, INF) - dp<<endl;

    return 0;
}