HDU 6070 Dirt Ratio（线段树）

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1473    Accepted Submission(s): 683
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 Xproblems 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 MyICPCLittle 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.

【题意】给你一个序列，问你对于任意 子序列，序列中数的种数/区间长度最小是多少。

【分析】二分答案midmid，检验是否存在一个区间满足cnt(l,r)/(r-l+1)≤mid，

 也就是cnt(l,r)+mid*l<=mid*(r+1)。从左往右枚举每个位置作为r，

 当r变化为r+1时，对cnt的影响是一段区间加1，线段树维护区间最小值即可。线段树维护的是当枚举到r时，每个区间内cnt(l,r)+mid*l的最小值，

 跟mid*(r+1)比较大小即可。这个题跟Codeforces Round #426 (Div. 2) D The Bakery很像，代码几乎一样，思想相同。

http://www.cnblogs.com/jianrenfang/p/7265602.html

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 6e4+5500;;
const int M = 160009;
const int mod = 1e9+7;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,k,ans;
int a[N],lazy[N*4];
int pre[N],pos[N];
double dp[N];
double mx[N*4];
void pushUp(int rt){
    mx[rt]=min(mx[rt<<1],mx[rt<<1|1]);
}
void pushDown(int rt){
    if(lazy[rt]){
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        mx[rt<<1]+=lazy[rt];
        mx[rt<<1|1]+=lazy[rt];
        lazy[rt]=0;
    }
}
void build(int l,int r,int rt,double x){
    lazy[rt]=0;
    if(l==r){
        mx[rt]=x*l;
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1,x);
    build(mid+1,r,rt<<1|1,x);
    pushUp(rt);
}
void upd(int L,int R,int l,int r,int x,int rt){
    if(L<=l&&r<=R){
        mx[rt]+=x;
        lazy[rt]+=x;
        return;
    }
    pushDown(rt);
    int mid=(l+r)>>1;
    if(L<=mid)upd(L,R,l,mid,x,rt<<1);
    if(R>mid) upd(L,R,mid+1,r,x,rt<<1|1);
    pushUp(rt);
}
double qry(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        return mx[rt];
    }
    pushDown(rt);
    double ret=100000000000;
    int mid=(l+r)>>1;
    if(L<=mid)ret=min(ret,qry(L,R,l,mid,rt<<1));
    if(R>mid)ret=min(ret,qry(L,R,mid+1,r,rt<<1|1));
    return ret;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        met(pos,0);
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            pre[i]=pos[a[i]];
            pos[a[i]]=i;
        }
        double l=0,r=1;
        for(int i=1;i<=20;i++){
            double mid = (l+r)/2;
            build(1,n,1,mid);
            bool ok=true;
            for(int j=1;j<=n;j++){
                upd(pre[j]+1,j,1,n,1,1);
                dp[j]=qry(1,j,1,n,1);
                if(dp[j]<=mid*(j+1)){
                    ok=false;break;
                }
            }
            if(!ok)r=mid;
            else l=mid;
        }
        printf("%.9f
",(l+r)/2);
    }
}