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

Posted 2020-10-08 Annetree的博客

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~

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 int a[2000005];
 7 int d[2000005];
 8 
 9 int main()
10 {
11     int x;
12     int n=0;
13     while(scanf("%d",&x)!=EOF)
14     {
15        if (x<0) continue;
16        if (x<10000) a[++n]=x;
17          else
18         for(int i=1;i<=5;i++) a[++n]=x-10000;
19     }
20     d[1]=a[1];  //初始化
21     int len=1;
22     for (int i=2;i<=n;i++)
23     {
24         if (a[i]>=d[len]) d[++len]=a[i];  //如果可以接在len后面就接上
25         else  //否则就找一个最该替换的替换掉
26         {
27             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
28             d[j]=a[i];
29         }
30     }
31     printf("%d\n",len);
32     return 0;
33 }