Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 1515    Accepted Submission(s): 698 Special Judge 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. Picture from MyICPC 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

题目大意：

让你找到一个区间：【L,R】，使得其中的数字的种类数/区间长度最小。

问这个比例最小是多少。

思路：

考虑分数规划：

①我们希望Val【L,R】/（R-L+1）最小。这里Val【L,R】表示区间中数字的种类数。

②我们不妨设定F（X）=Val【L,R】-X*（R-L+1）；

③那么如果有F（X）<=0，那么一定有Val【L,R】/（R-L+1）<=X；

④所以我们这个最小的值可以通过二分来求出来。

那么我们如何check呢？

我们再转化一下不等式变成：Val【L,R】+mid*L<=（R+1）*mid.（这里X==mid）；

那么我们O（n）枚举R，之前我们可以利用线段树来维护mid*L+Val【L,R】的最小值，那么如果查询Query（1，i-1）能够<=（R+1）*mid的话，就说明可以继续向下二分。

否则向上。

我们线段树更新mid*L很简单，一路update（i，i，mid*i）即可，那么Val【L,R】怎样处理呢？其实也不难，我们维护一个数组last【i】，表示数字i出现的最后的位子，那么对于当前数i，其影响的区间Val的值的范围就是从【last【a【i】】+1，i】，那么再update一下（last【a【i】】+1，i，1）即可。

尽量去for二分。

Ac代码：

#include<stdio.h> #include<iostream> #include<string.h> using namespace std; #define lson l,m,rt*2 #define rson m+1,r,rt*2+1 double tree[350050*8]; double flag[350050*8]; int last[350050]; int a[350050]; void build(int l,int r,int rt){ flag[rt]=0; if(l==r) { tree[rt]=0; flag[rt]=0; return; } int m=(l+r)>>1; build(lson); build(rson); } void pushdown(int l,int r,int rt)//向下维护树内数据 { if(flag[rt])//如果贪婪标记不是0（说明需要向下进行覆盖区间（或点）的值） { int m=(l+r)/2; flag[rt*2]+=flag[rt]; flag[rt*2+1]+=flag[rt]; tree[rt*2]+=flag[rt];//千万理解如何覆盖的区间值（对应线段树的图片理解（m-l）+1）是什么意识. tree[rt*2+1]+=flag[rt]; flag[rt]=0; } } void pushup(int rt) { tree[rt]=max(tree[rt<<1],tree[rt<<1|1]); } double Query(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R) { return tree[rt]; } else { pushdown(l,r,rt); int m=(l+r)>>1; double ans=-10000000000000; if(L<=m) { ans=max(ans,Query(L,R,lson)); } if(m<R) { ans=max(ans,Query(L,R,rson)); } pushup(rt); return ans; } } void update(int L,int R,double c,int l,int r,int rt) { if(L<=l&&r<=R)//覆盖的是区间~ { tree[rt]+=c;//覆盖当前点的值 flag[rt]+=c;//同时懒惰标记~！ return ; } pushdown(l,r,rt); int m=(l+r)/2; if(L<=m) { update(L,R,c,lson); } if(m<R) { update(L,R,c,rson); } pushup(rt); } int Slove(int n,double mid) { build(1,n,1); for(int i=1;i<=n;i++)last[a[i]]=0; for(int i=1;i<=n;i++) { update(i,i,-mid*i,1,n,1); update(last[a[i]]+1,i,-1,1,n,1); last[a[i]]=i; if(i==1)continue; double Q=-Query(1,i-1,1,n,1); if(Q<=(i+1)*mid)return 1; } return 0; } int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d",&a[i]); double l=0; double r=1; double ans; for(int i=0;i<20;i++) { double mid=(l+r)/2; if(Slove(n,mid)==1) { ans=mid; r=mid; } else l=mid; } printf("%.10f\n",ans); } }