2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 L. The Heaviest Non-decreasing Subsequence Problem （LIS）

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原题：

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

思路：

将权值为5的存入五次，权值为1的存入一次，权值为0的不存，LIS即可。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <string>#include <stack>#include <queue>#include <cmath>#include <map>#include <utility>using namespace std;typedef long long ll;const int INF = 0x3f3f3f3f;const int MAXN = 5 * 1e5 + 10;vector <int > v;int b[MAXN];int LIS (int n){    int cnt = 0;    b[0] = -1;    for (int i = 0; i < n; i++)    {        if (v[i] >= b[cnt]) b[++cnt] = v[i];        else        {            int pos = upper_bound(b + 1, b + 1 + cnt, v[i])-b;            b[pos] = v[i];        }    }    return cnt;}int main(){    v.clear();    int num;    while(scanf("%d",&num)!=EOF)    {        if(num<0)            continue;        if(num<10000)        {            v.push_back(num);            continue;        }        else            for(int i=0;i<5;i++)            v.push_back(num-10000);    }//    for(int i=0;i<v.size();i++)//        cout<<v[i]<<" ";//    cout<<endl;    int len=v.size();    int res=LIS(len);    printf("%d\n",res);    return 0;}

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