Dirt Ratio

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题目链接

题意

给$n$个数，计算$(l,r)$中不同数字数量/$(r-l+1)$.即 的最小值。

思路

二分答案，每次去看$Mid$值是否合法。我们将上面式子变形为

我们枚举$r$，在线段树上维护到当前$r$每一$l$的.动态更新会产生影响的。

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50
#define lson rt << 1
#define rson rt << 1 | 1
#define Lson L, mid, lson
#define Rson mid + 1, R, rson
const int N = 6e4 + 5;
double lazy[N << 2];
double val[N << 2];
double Mid;
void push_up(int rt)
{
    //sum[rt] = sum[lson] + sum[rson];
    val[rt] = min(val[lson], val[rson]);
}

void build(int L, int R, int rt)
{
    //sum[rt] = 0;
    lazy[rt] = 0;
    val[rt] = inf;
    if (L == R)
    {
        val[rt] = L * Mid;
        return;
    }
    int mid = (L + R) >> 1;
    build(Lson);
    build(Rson);
    push_up(rt);
}

void push_down(int len, int rt)
{
    if (lazy[rt])
    {
//        sum[lson] = sum[lson] + (len - (len >> 1)) * lazy[rt];
//        sum[rson] = sum[rson] + (len >> 1) * lazy[rt];
        val[lson] = val[lson] + lazy[rt];
        val[rson] = val[rson] + lazy[rt];
        lazy[lson] += lazy[rt];
        lazy[rson] += lazy[rt];
        lazy[rt] = 0;
    }
}

void update(int l, int r, int v, int L, int R, int rt)
{
    
    if (l <= L && r >= R)
    {
        int len = R - L + 1;
        lazy[rt] += v;
//        sum[rt] += v * len;
        val[rt] += v;
        return;
    }
    /*
    if (L == R)
    {
        val[rt] += v;
        return;
    }
    */
    int mid = (L + R) >> 1;
    push_down(R - L + 1, rt);
    if (l <= mid)
        update(l, r, v, Lson);
    if (r > mid)
        update(l, r, v, Rson);
    push_up(rt);
}

double query(int l, int r, int L, int R, int rt)
{
    if (l <= L && r >= R)
        return val[rt];
    double ans = (1ll << 50) * 1.0;
    push_down(R - L + 1, rt);
    int mid = (L + R) >> 1;
    if (l <= mid)
        ans = min(ans, query(l, r, Lson));
    if (r > mid)
        ans = min(ans, query(l, r, Rson));
    return ans;
}

int pre[N];
int pos[N];
int a[N], n;

bool check()
{
    //dbg(Mid);
    build(1, n, 1);
    for (int i = 1; i <= n; i++)
    {
    //    dbg(i);
        update(pre[i] + 1, i, 1, 1, n, 1);
    //    dbg(query(1, i, 1, n, 1));
        if (query(1, i, 1, n, 1) <= Mid * 1.0 * (i + 1))
            return true;
    }
    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for (int i = 1; i <= n; i++)
            pos[i] = 0;
        for (int i = 1; i <= n; i++)
        {
            pre[i] = pos[a[i]];
            pos[a[i]] = i;
        }
        double l = 0, r = n + 1, eps = 1e-6;
        while (r - l >= eps)
        {
            Mid = (r + l) / 2.0;
            if (check())
                r = Mid;
            else
                l = Mid;
        }
        printf("%.5f\n", l);
    }
    return 0;
}
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