P3299 [SDOI2013]保护出题人

发表于 | 阅读数

简单题意，不说。

可以看成$(x_i+i\times d,sum_i),(j\times d,sum_{j-1})$两点的斜率
询问不单调，三分凸包最大值。

代码 
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#include <bits/stdc++.h>
                                                           using namespace std;
typedef long long ll;
typedef double db;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
ll sum[N], a[N], x[N];
db ans = 0;
ll d;
db f(int i, int j)
{
    return 1.00 * (sum[i] - sum[j - 1]) / (x[i] + (i - j) * d);
}
db slope(int i, int j)
{
    return 1.00 * (sum[i - 1] - sum[j - 1]) / (i * d - j * d);
}
int q[N];
int main()
{
    int n = read();
    d = read();
    for (int i = 1; i <= n; i++)
    {
        a[i] = read();
        x[i] = read();
    }
    for (int i = 1; i <= n; i++)
        sum[i] = a[i] + sum[i - 1];
    int head = 1, tail = 0;
    for (int i = 1; i <= n; i++)
    {
        while (head < tail && slope(q[tail], i) < slope(q[tail - 1], q[tail]))
            tail--;
        q[++tail] = i;
        int l = 1, r = tail;
        while (l != r)
        {
            int x = (r - l + 1) / 3;
            int mid1 = l + x - 1, mid2 = l + 2 * x - 1;
            if (mid1 == mid2)
                break;
            if (f(i, q[mid1]) < f(i, q[mid2]))
                l = mid1 + 1;
            else
                r = mid2 - 1;
        }

        ans += max(f(i, q[l]), f(i, q[r]));
    }
    printf("%.0lf\n", ans);
}