The Heaviest Non-decreasing Subsequence Problem（2017南宁网络赛）

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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 .

(2) If is is greater than or equal to , then its weight is . Furthermore, the real integer value of $s_{i}$ is $s_{i}-10000$ . For example, if $s_{i}$ is , then is is reset to and its weight is .

(3) Otherwise, its weight is .

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 , 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:

The heaviest non-decreasing subsequence of the sequence is with the total weight being . 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

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80 75 73 93 73 73 10101 97 -1 -1 114 -1 10113 118

样例输出

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1
14

//题意：给定一个数组，现规定，里面的数值为负的话，它的weight是0，若大于等于0且小于10000的话，其weight为1，若大于等于10000，它的weight为5，且它的值是原来的值减10000。求这个数组的子串中weight和的最大值。

//思路：开一个新数组，遍历给定的数组，值为负的直接舍去，若值是小于10000的非负数，就依次存到一个新数组，若值大于10000，将这个值减去10000后，放入新数组中，放5次，因为它的weught=5，然后就得到一个新数组，求这个新数组的非递减最长子序列的长度即可。

复制代码

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

const int MAX=2e5+100;
int num[MAX];
int a[MAX];
int dp[MAX];
int cnt;

int bsearch(int len,int x)
{
    int l=0,r=len-1;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(x>=dp[mid-1]&&x<dp[mid])return mid;
        else if(x<dp[mid-1])r=mid-1;
        else l=mid+1;
    }
    return l;
}
//最长非递减子序列
int LIS()
{
    dp[0]=a[0];
    int len=1;
    int j;
    for(int i=1; i<cnt; i++)
    {
        if(a[i]<dp[0])j=0;
        else if(a[i]>=dp[len-1])j=len++;
        else j=bsearch(len,a[i]);
        dp[j]=a[i];
    }
    return len;
}

int main()
{
    memset(num,0,sizeof(num));
    memset(dp,0,sizeof(dp));
    memset(a,0,sizeof(a));
    int nn=0;
    while(scanf("%d",&num[nn])!=EOF)
    {
        nn++;
    }
    cnt=0;
    for(int i=0;i<nn;i++)
    {
        if(num[i]<0)
            continue;
        if(num[i]>=10000)
        {
            num[i]-=10000;
            for(int j=0;j<5;j++)
            {
                a[cnt++]=num[i];
            }
        }
        else
        {
            a[cnt++]=num[i];
        }
    }
    int ans=LIS();
    printf("%dn",ans);
    return 0;
}

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#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

const int MAX=2e5+100;
int num[MAX];
int a[MAX];
int dp[MAX];
int cnt;

int bsearch(int len,int x)
{
    int l=0,r=len-1;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(x>=dp[mid-1]&&x<dp[mid])return mid;
        else if(x<dp[mid-1])r=mid-1;
        else l=mid+1;
    }
    return l;
}
//最长非递减子序列
int LIS()
{
    dp[0]=a[0];
    int len=1;
    int j;
    for(int i=1; i<cnt; i++)
    {
        if(a[i]<dp[0])j=0;
        else if(a[i]>=dp[len-1])j=len++;
        else j=bsearch(len,a[i]);
        dp[j]=a[i];
    }
    return len;
}

int main()
{
    memset(num,0,sizeof(num));
    memset(dp,0,sizeof(dp));
    memset(a,0,sizeof(a));
    int nn=0;
    while(scanf("%d",&num[nn])!=EOF)
    {
        nn++;
    }
    cnt=0;
    for(int i=0;i<nn;i++)
    {
        if(num[i]<0)
            continue;
        if(num[i]>=10000)
        {
            num[i]-=10000;
            for(int j=0;j<5;j++)
            {
                a[cnt++]=num[i];
            }
        }
        else
        {
            a[cnt++]=num[i];
        }
    }
    int ans=LIS();
    printf("%dn",ans);
    return 0;
}