2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 The Heaviest Non-decreasing Subsequence Problem 最长不下降序列

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_{2}\ ...\ <i_{k}i​1​​<i​2​​ ... <i​k​​, such that, for all $1\le j<k$, we have s_{ij}<s_{ij+1}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

    题目是说找出一个值最大的序列，而这个序列的值的定义方式是对于负数，值为0。对于正数，如果小于10000，那么值就是1，否则为5。另外，如果大于10000，那么这个数要减去10000才是他真正在数列中的数（比如10101，在数列中也将其视为101）。然后将其排列出一个最长的序列，要求这个序列的值最大，且序列为不下降的序列。

    处理的时候，对于负数，因为没有值，所以直接舍去不读入就可以了。对于正数，如果小于10000就正常读入，如果大于10000的话，先将这个数减去10000存入，因为这个数的值为5，所以再复制4次，存入这个数5次进去，这样就相当于处理了权值为5的问题，然后读入完之后求出最长不下降子序列就可以了。、

    （PS：这个题输入是一组数据，比赛的时候我当成多组数据写了，然后写了很多getchar判断空格回车结束，结果TLE了两发改成一组数据才A了....）

    下面AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[1000005];int b[1000005],c[1000005];int fin(int *t,int len,int n){    int left=0,right=len,mid=(left+right)/2;    while(left<=right)    {        if(n>=t[mid])            left=mid+1;        else if(n<t[mid])            right=mid-1;        else            return mid;        mid=(left+right)/2;    }    return left;}int main(){    int n;    int i,j;    int k;    int m;    int t;    int len;    char g;    i=0;    while(scanf("%d",&m)!=EOF)    {        a[i]=m;        if(a[i]<0)        {            i--;        }        else if(a[i]>10000)        {            a[i]-=10000;            k=a[i];            for(j=1;j<=4;j++)            {                i++;                a[i]=k;            }        }        i++;    }    n=i;    /*for(i=0;i<n;i++)    {        cout<<"a["<<i<<"] = "<<a[i]<<endl;    }*/    b[0]=1;    c[0]=-1;    c[1]=a[0];    len=1;    for(i=1;i<n;i++)    {        j=fin(c,len,a[i]);        c[j]=a[i];        if(j>len)            len=j;    }    cout<<len<<endl;    return 0;}