2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 L

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

Let $S$ be a sequence of integers $s_{1}$ , $s_{2}$ , $. . .$ , $s_{n}$ 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 $s_{i}$ is $s_{i}-10000$ . For example, if $s_{i}$ 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 $s_{i1}$ , $s_{i2}$ , $. . .$ , $s_{ik}$ , with $i_{1}<i_{2}\ ...\ <i_{k}$ , such that, for all $\leq j<k$ , we have $s_{ij}<s_{ij+1}$ .

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 $2*10^{5}$

Input Format

A list of integers separated by blanks: $s_{1}$ , $s_{2}$ , $. . .$ , $s_{n}$

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

#include<cstdio>#include<algorithm>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

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