HDU 6070 Dirt Ratio（线段树）

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 Y. 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 ), 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 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\n",(l+r)/2);
    }
}