计蒜客 The Heaviest Non-decreasing Subsequence Problem（最大权值和非递减子序列）

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

先把数值转换，然后把weight转换为连续出现次数，套用最长非递减子序列模板。

#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>using namespace std;const int maxn=1000010;int data[maxn];int maxv[maxn];int BinaryResearch(int x,int len){int mid,low=1,high=len;while(low<=high){mid=(low+high)>>1;if(maxv[mid]<=x){low=mid+1;}else high=mid-1;}return low;}int lis(int n){int i,len=1;maxv[1]=data[0];for(i=1;i<n;i++){if(data[i]>=maxv[len]){maxv[++len]=data[i];}else{int pos=BinaryResearch(data[i],len);maxv[pos]=data[i];}}return len;}int main(){memset(data,0,sizeof(data));memset(maxv,0,sizeof(maxv));int len=0;int x;while(scanf("%d",&x)!=EOF){if(x<0)continue;else if(x>=10000){x=x-10000;data[len++]=x;data[len++]=x;data[len++]=x;data[len++]=x;data[len++]=x;}else{data[len++]=x;}}printf("%d\n",lis(len));}