2017ICPC南宁赛区网络赛 The Heaviest Non-decreasing Subsequence Problem （最长不下降子序列）

Let SSS be a sequence of integers s1s_{1}s​1​​, s2s_{2}s​2​​, ........., sns_{n}s​n​​ Each integer is is associated with a weight by the following rules:

(1) If is is negative, then its weight is 000.

(2) If is is greater than or equal to 100001000010000, then its weight is 555. Furthermore, the real integer value of sis_{i}s​i​​ is si−10000s_{i}-10000s​i​​−10000 . For example, if sis_{i}s​i​​ is 101011010110101, then is is reset to 101101101 and its weight is 555.

(3) Otherwise, its weight is 111.

A non-decreasing subsequence of SSS is a subsequence si1s_{i1}s​i1​​, si2s_{i2}s​i2​​, ........., siks_{ik}s​ik​​, with i1<i2 ... <iki_{1}<i_{2}\ ...\ <i_{k}i​1​​<i​2​​ ... <i​k​​, such that, for all 1≤j<k1 \leq j<k1≤j<k, we have sij<sij+1s_{ij}<s_{ij+1}s​ij​​<s​ij+1​​.

A heaviest non-decreasing subsequence of SSS 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:

808080 757575 737373 939393 737373 737373 101011010110101 979797 −1-1−1 −1-1−1 114114114 −1-1−1 101131011310113 118118118

The heaviest non-decreasing subsequence of the sequence is <73,73,73,101,113,118><73, 73, 73, 101, 113, 118><73,73,73,101,113,118> with the total weight being 1+1+1+5+5+1=141+1+1+5+5+1 = 141+1+1+5+5+1=14. Therefore, your program should output 141414 in this example.

We guarantee that the length of the sequence does not exceed 2∗1052*10^{5}2∗10​5​​

Input Format

A list of integers separated by blanks:s1s_{1}s​1​​, s2s_{2}s​2​​,.........,sns_{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

超过一万的，权重是5，所以把它变成5个就好啦。这样的话问题就转化成了最长不下降子序列~

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int a[2000005];
int d[2000005];

int main()
{
int x;
int n=0;
while(scanf("%d",&x)!=EOF)
{
if (x<0) continue;
if (x<10000) a[++n]=x;
else
for(int i=1;i<=5;i++) a[++n]=x-10000;
}
d[1]=a[1];  //初始化
int len=1;
for (int i=2;i<=n;i++)
{
if (a[i]>=d[len]) d[++len]=a[i];  //如果可以接在len后面就接上
else  //否则就找一个最该替换的替换掉
{
int j=upper_bound(d+1,d+len+1,a[i])-d;  //找到第一个大于它的d的下标
d[j]=a[i];
}
}
printf("%d\n",len);
return 0;
}