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The Heaviest Non-decreasing Subsequence Problem
阅读量:5240 次
发布时间:2019-06-14

本文共 2183 字,大约阅读时间需要 7 分钟。

Let S be a sequence of integers s1s_{1}s1​​, s2s_{2}s2​​, ........., sns_{n}sn​​ 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 sis_{i}si​​ is si−10000s_{i}-10000si​​10000 . For example, if sis_{i}si​​ is 101011010110101, then is is reset to 101101101 and its weight is 555.

(3) Otherwise, its weight is 1.

A non-decreasing subsequence of S is a subsequence si1s_{i1}si1​​, si2s_{i2}si2​​, ........., siks_{ik}sik​​, with i1<i2 ... <iki_{1}<i_{2}\ ...\ <i_{k}i1​​<i2​​ ... <ik​​, such that, for all 1≤j<k1 \leq j<k1j<k, we have sij<sij+1s_{ij}<s_{ij+1}sij​​<sij+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∗1052*10^{5}2105​​

Input Format

A list of integers separated by blanks:s1s_{1}s1​​, s2s_{2}s2​​,.........,sns_{n}sn​​

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

求最长曾序列,dp

1 #include 
2 #define ll long long 3 using namespace std; 4 const int N = 2e5+10; 5 int a[N*5], dp[N*5]; 6 int main() { 7 int n, num = 0; 8 while(scanf("%d", &n) != EOF) { 9 if(n >= 10000) {10 for(int i = 0; i < 5; i ++) a[num++] = n - 10000;11 } else if(n >= 0) a[num++] = n;12 }13 for(int i = 0; i <= N*5; i ++) dp[i] = 1000000;14 for(int i = 0; i < num; i ++) {15 *upper_bound(dp, dp+num, a[i]) = a[i];16 }17 cout << lower_bound(dp,dp+num,1000000)-dp << endl ;18 return 0;19 }

 

转载于:https://www.cnblogs.com/xingkongyihao/p/7588594.html

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