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

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

题意：带权重的最长不下降子序列……

解题思路：由于权重很小，直接把数重复多遍即可。例如 1 权重为 5 那么就变成 1 1 1 1 1 然后再求最长不下降子序列。要用nlogn的算法……

#include<iostream>#include<algorithm>#include<math.h>using namespace std;typedef long long int ll;const int INF=(1<<30);int a[2000005];int d[2000005];int main(){        int n=1;        int temp;        while(~scanf("%d",&temp)){        if(temp>=10000){            temp-=10000;            a[n++]=temp;            a[n++]=temp;            a[n++]=temp;            a[n++]=temp;            a[n++]=temp;                    }        else{            if(temp<0){                ;//负数直接去掉这个数就好了            }                        else{                a[n++]=temp;            }                    }            }        d[1]=a[1];     int len=1;    for (int i=2;i<=n;i++)    {        if (a[i]>=d[len]) d[++len]=a[i];        else        {            int j=upper_bound(d+1,d+len+1,a[i])-d;            d[j]=a[i];         }    }        printf("%d\n",len);            return 0;}