HDU 6070 Dirt Ratio【分数规划】【线段树】

题目来戳呀

Problem Description

In ACM/ICPC contest, the ”Dirt Ratio” of a team is calculated in the following way. First let’s ignore all the problems the team didn’t pass, assume the team passed X problems during the contest, and submitted Y times for these problems, then the ”Dirt Ratio” is measured as XY. If the ”Dirt Ratio” of a team is too low, the team tends to cause more penalty, which is not a good performance.

Little Q is a coach, he is now staring at the submission list of a team. You can assume all the problems occurred in the list was solved by the team during the contest. Little Q calculated the team’s low ”Dirt Ratio”, felt very angry. He wants to have a talk with them. To make the problem more serious, he wants to choose a continuous subsequence of the list, and then calculate the ”Dirt Ratio” just based on that subsequence.

Please write a program to find such subsequence having the lowest ”Dirt Ratio”.

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there is an integer n(1≤n≤60000) in the first line, denoting the length of the submission list.

In the next line, there are n positive integers a1,a2,…,an(1≤ai≤n), denoting the problem ID of each submission.

Output

For each test case, print a single line containing a floating number, denoting the lowest ”Dirt Ratio”. The answer must be printed with an absolute error not greater than 10−4.

Sample Input

1

5

1 2 1 2 3

Sample Output

0.5000000000

Hint

For every problem, you can assume its final submission is accepted.

Source

2017 Multi-University Training Contest - Team 4

题意

求区间数字种类的个数/区间长度的最大值。

想法

假设结果为mid，求mid=sizer−l+1的最值，可以转化为求size+l*mid=(r+1)*mid。用线段树保存size(l,r)+mid*l，枚举mid，二分来自做。

#include<bits/stdc++.h>
using namespace std;
const int maxn=60010;
//const double acc=0.0001;
int num[maxn<<2],pos[maxn],a[maxn];
double v[maxn<<2];
double MID,tmp,L,R;
void build(int x,int l,int r)
{//建树
v[x]=MID*l;//l是每次枚举的左端点，相当于mid*l是已知的
num[x]=0;
if(l==r)
return;
int mid=(l+r)>>1;
build(x<<1,l,mid);
build(x<<1|1,mid+1,r);
}
void add(int x,int cnt)
{//单点更新
v[x]+=cnt;
num[x]+=cnt;
}
void pushdown(int x)
{//对子节点进行延迟标记
if(num[x])
{
add(x<<1,num[x]);
add(x<<1|1,num[x]);
num[x]=0;
}
}
void update(int x,int l,int r,int pre,int right)
{//更新每个点
if(pre<=l&&r<=right)
{
add(x,1);//这道题在区间内未出现过 所以贡献一次
return;
}
pushdown(x);
int mid=(l+r)>>1;
if(pre<=mid)
update(x<<1,l,mid,pre,right);
if((mid+1)<=right)
update(x<<1|1,mid+1,r,pre,right);
v[x]=min(v[x<<1],v[x<<1|1]);
}
void query(int x,int l,int r,int right)
{
//size-mid*l<=(r+1)*mid
if(r<=right)
{

if(tmp>v[x])
tmp=v[x];
return;
}
int mid=(l+r)>>1;
query(x<<1,l,mid,right);
if((mid+1)<=right)
query(x<<1|1,mid+1,r,right);
}
int main()
{
int T,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int k=1; k<=n; ++k)
scanf("%d",&a[k]);
L=0.0,R=1.0;
for(int j=1; j<=24; ++j)
{
MID=(L+R)/2;
build(1,1,n);
int i;
for(i=1; i<=n; ++i)
pos[i]=0;///有问题 下标
for(i=1; i<=n; ++i)
{
tmp=0x3f3f3f3f;//初始化最大值
update(1,1,n,pos[a[i]]+1,i);
query(1,1,n,i);
if(tmp-MID*(i+1)<=0)//这里就是公式的转化，size+MID*l-MID*(r+1)
break;
pos[a[i]]=i;
}

if(i<=n)R=MID;
else L=MID;
}

printf("%.10f",(L+R)/2);
}
return 0;
}

ps：比较匆忙，比赛完再补一下。而且代码还不是优美的，丑哭QAQ