2017 ACM-ICPC 亚洲区（南宁赛区）网络赛：L. The Heaviest Non-decreasing Subsequence Problem

Let $S$ be a sequence of integers s_{1}s​1​​, s_{2}s​2​​, $...$, s_{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 s_{i}s​i​​ is s_{i}-10000s​i​​−10000 . For example, if s_{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 s_{i1}s​i1​​, s_{i2}s​i2​​, $...$, s_{ik}s​ik​​, with i_{1}<i_{2}\ ...\ <i_{k}i​1​​<i​2​​ ... <i​k​​, such that, for all $1\le j<k$, we have s_{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*10^{5}2∗10​5​​

Input Format

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

思路：将价值为5的数变为5个价值为1的数，再求最长上升子序列即可。

#include<bits/stdc++.h>using namespace std;const int MAX = 2e6;int dp[MAX],a[MAX];int main(){int n;char ch;while(scanf("%d%c",&n,&ch)!=EOF){if(ch == '\n'){            if(n<0)puts("0");            else if(n>=10000)puts("5");            else puts("1");continue;}int t=1;if(n >= 10000){a[t++] = n-10000;            a[t++] = n-10000;            a[t++] = n-10000;            a[t++] = n-10000;            a[t++] = n-10000;}else if(n>=0)a[t++] = n;while(scanf("%d%c",&n,&ch)!=EOF){if(n >= 10000){a[t++] = n-10000;a[t++] = n-10000;a[t++] = n-10000;a[t++] = n-10000;a[t++] = n-10000;}else if(n>=0)a[t++] = n;if(ch == '\n')break;}int len = 0;for(int i = 1; i < t; i++){if(len==0||a[i] >=dp[len])dp[++len] = a[i];else{int cnt = upper_bound(dp,dp+1+len,a[i]) - dp;dp[cnt] = a[i];}}printf("%d\n",len);}return 0;}