线段树+二分——HDU6070

题解：

题目让我们求type(l,r)/(r-l+1)的最小，显然可以用01分数规划，来二分求最小答案。所以我们从左往右移动端点，来统计种类。线段树来维护type(l,r)+l*mid的值。

#include 
using namespace std;
const int N = 1e6 + 7;
const double eps = 1e-7;
double sum[N];
int lazy[N];
int pre[60007];
int a[60007];
int n;
void build(int l, int r, int rt, double mi) {
    lazy[rt] = 0; //初始种类数为零
    if (l == r) {
        sum[rt] = l * mi; // l*mid
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1, mi);
    build(mid + 1, r, rt << 1 | 1, mi);
    sum[rt] = min(sum[rt<<1], sum[rt<<1|1]);
}
void pushdown(int rt) {
    lazy[rt << 1] += lazy[rt];
    lazy[rt << 1|1] += lazy[rt];
    sum[rt << 1] += lazy[rt];
    sum[rt << 1|1] += lazy[rt];
    lazy[rt] = 0;
}
void update(int l, int r, int rt, int L, int R) {
    if (L <= l && r <= R) {
        sum[rt] += 1;
        lazy[rt] += 1;
        return;
    }
    if (lazy[rt]) pushdown(rt);
    int mid = (l + r) >> 1;
    if (L <= mid) update(l, mid, rt<<1, L, R);
    if (R > mid) update(mid+1, r, rt<<1|1, L, R);
    sum[rt] = min(sum[rt<<1], sum[rt<<1|1]);
}
double query(int l, int r, int rt, int L, int R) {
    if (L <= l && r <= R) {
        return sum[rt];
    }
    if (lazy[rt]) pushdown(rt);
    int mid = (l + r) >> 1;
    double tmp = 1e9;
    if (L <= mid) tmp = min(tmp, query(l, mid, rt<<1, L, R));
    if (R > mid) tmp = min(tmp, query(mid+1, r, rt<<1|1, L, R));
    return tmp;
}

int check(double mid) {
    build(1, n, 1, mid);
    memset(pre, 0, sizeof(pre));
    for (int i = 1; i <= n; ++i) {
        update(1, n, 1, pre[a[i]]+1, i);
        if (query(1, n, 1, 1, i) - mid*(i+1) < eps) return 1;
        pre[a[i]] = i;
    }
    return 0;
}
int main() {
    int t;
    scanf ("%d", &t);
    while (t--) {
        scanf ("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf ("%d", &a[i]);
        }
        double l = 0, r = 1;
        while (r - l > eps) {
            double mid = (l + r)/2.0;
            if (check(mid)) r = mid;
            else l = mid;
        }
        printf ("%0.5f\n", l);
    }
    return 0;
}