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

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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 $0$ .

(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

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

分析：人生的第一个树状数组优化dp题目，尴尬的是队友会dp，我会树状数组，很遗憾，没做出来。

这里介绍两种思路。

思路见代码 :

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;

int tree[300005];
struct Node {
	int a[300005];
	int val[300005];
}data;
int dp[300005];
void add(int x,int num)
{
    for(int i=x;i<=300005;i+=i&-i)
    	if(num>tree[i])
        	tree[i]=num;
}
int read(int x)
{
    int num=0;
    for(int i=x;i>0;i-=i&-i)
    	if(num<tree[i])
    		num=tree[i];
    return num;
}
int main()
{
    int n=1;
    while(~scanf("%d",&data.a[n]))
	{
		if(data.a[n]<0)	//负数可以不存 
			continue;
		else if(data.a[n]>=10000)
		{
			data.a[n]-=10000;	//大于10000的数，real值是原数减10000 
			data.val[n]=5;
		}
		else
			data.val[n]=1;
		n++;
	}
	dp[1]=data.val[1];
	add(data.a[1],dp[1]);
	int ans=-1;
	for(int i=2;i<n;i++)
	{//为什么说是树状数组优化dp，因为这里本来是又一层循环，用来查找在满足data.a[j]<=data.a[i]的情况下，
	//寻找最大的dp[j];	所以咱们以 data.a[i]为区间坐标，求区间最值，从0~data.a[i]中找到的最大的dp[j],
	//一定满足 data.a[j]<=data.a[i]
		int j=read(data.a[i]);
		dp[i]=max(dp[i],j+data.val[i]);
		add(data.a[i],dp[i]);
		ans=max(ans,dp[i]);
	}
	printf("%d\n",ans);
    return 0;
}

第二种：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;

int a[1000005];
int dp[1000005];
int main()
{
    int n;
    n=1; 
    while(~scanf("%d",&a[n]))
    {//既然只有val==1，5的时候有用，而且求的是不下降的heavies序列。那么不妨将这些val=5的
	//拆分成5个val=1的，然后直接求解最大不下降子序列。 
    	if(a[n]<0)
    		continue;
    	else if(a[n]>=10000)
    	{
    		a[n]-=10000;
    		int tem=a[n];
    		n++;
    		for(int i=0;i<4;i++)
				a[n++]=tem;
		}
    	else
    		n++;
	}
    int len=1;
    memset(dp,0,sizeof(dp));
    dp[1]=a[1];
    for(int i=2;i<n;i++)
    {
        if(a[i]>=dp[len])
            dp[++len]=a[i];
        else
        {
            int add = lower_bound(dp+1,dp+len+1,a[i]) - dp;
            while(dp[add]<=a[i])
            	add++;
            dp[add] = a[i];
        }
    }
    if(n==1 && a[n]<0)//注意边界情况处理 
		printf("0\n");
	else
		printf("%d\n",len);
    
    return 0;
}