HDU 6070 - Dirt Ratio | 2017 Multi-University Training Contest 4

比赛时会错题意+不知道怎么线段树维护分数- -

思路来自题解

/*
HDU 6070 - Dirt Ratio [ 二分,线段树 ]  |  2017 Multi-University Training Contest 4
题意：
	给出 a[N];
	设 size(l,r)为区间(l,r)不同数字的个数，求 size(l,r)/(r-l+1) 的最小值
	限制： N <= 6e5, a[i] <= 6e5
分析：
	二分答案 mid
	则判定条件为是否存在 size(l,r)/(r-l+1) <= mid 
	变换一下：  size(l,r) + mid*l <= mid * (r+1)
	将左式存入线段树中，枚举 r，对某段 l 进行更新（last[a[r]+1] 到 r），更新操作为值+1
		再对每个 r 判断一下上式是否成立
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 6e4+5;
const double eps = 1e-5;
const double INF = 1e18;
namespace SegT {
    double val[N];
    double Min[N<<2]; int add[N<<2];
    void up(int x) {
        Min[x] = min(Min[x<<1], Min[x<<1|1]);
    }
    void down(int x) {
        if (add[x]) {
            add[x<<1] += add[x];
            Min[x<<1] += add[x];
            add[x<<1|1] += add[x];
            Min[x<<1|1] += add[x];
            add[x] = 0;
        }
    }
    void build(int l, int r, int x) {
        add[x] = 0;
        if (l == r) {
            Min[x] = val[l]; return;
        }
        int mid = (l+r) >> 1;
        build(l, mid, x<<1);
        build(mid+1, r, x<<1|1);
        up(x);
    }
    void change(int L, int R, int num, int l, int r, int x) {
        if (L <= l && r <= R) {
            add[x] += num;
            Min[x] += num;
            return;
        }
        down(x);
        int mid = (l+r) >> 1;
        if (L <= mid) change(L, R, num, l, mid, x<<1);
        if (mid < R) change(L, R, num, mid+1, r, x<<1|1);
        up(x);
    }
    double query(int L, int R, int l, int r, int x) {
        if (L <= l && r <= R) return Min[x];
        down(x);
        int mid = (l+r) >> 1;
        double res = INF;
        if (L <= mid) res = min(res, query(L, R, l, mid, x<<1));
        if (R > mid) res = min(res, query(L, R, mid+1, r, x<<1|1));
        return res;
    }
}
int t, n, a[N];
int last[N];
bool solve(double mid)
{
    for (int i = 1; i <= n; i++)
        SegT::val[i] = i*mid;
    SegT::build(1, n, 1);
    memset(last, 0, sizeof(last));
    for (int i = 1; i <= n; i++)
    {
        SegT::change(last[a[i]]+1, i, 1, 1, n, 1);
        last[a[i]] = i;
        double res = SegT::query(1, i, 1, n, 1);
        if (res < (i+1)*mid - eps) return 1;
    }
    return 0;
}
double BinaryFind(double l, double r)
{
    double mid;
    while ((r-l) > eps) {
        mid = (l+r) / 2;
        if (solve(mid)) r = mid;
        else l = mid;
    }
    return mid;
}
int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        printf("%.9f\n", BinaryFind(0, 1));
    }
}