2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 L

 The Heaviest Non-decreasing Subsequence Problem（最长不下降子序列变形）

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​1​​<i​2​​ ... <i​k​​, such that, for all 1 \leq j1≤j<k, we have s_{ij}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

#include
#include
using namespace std;
const int maxn = 1e6+5;//范围有毒，不要信题目的保证。尽量开大一点，反正不要钱。
int a[maxn], d[maxn];
int main()
{
	int x,n=0,len=1;
	while (~scanf("%d", &x))//存序列
	{
		if (x < 0) continue;
		else if (x < 10000)
		{
			a[n++] = x;
		}
		else
		{
			for (int j = 0; j < 5; j++)
			{
				a[n++] = x - 10000;
			}
		}
	}
		d[1] = a[0];
		for (int i = 1; i < n; i++)//数组d保存长度为其下标的子序列的最小值
		{
			if (d[len] <= a[i])
			{
				len++;
				d[len] = a[i];
			}
			else
			{
				int j = upper_bound(d + 1, d + 1 + len, a[i])-d;//找到第一个大于a[i]的d[j]
				d[j] = a[i];
			}
		}
	printf("%d\n", len);//权值都为1，权值之和等于长度
	return 0;
}

转载地址：http://blog.csdn.net/lzc504603913/article/details/78078245