[SDOI 2016]征途

时间：2022-03-02 04:56:06

Description

题库链接

将一个长度为 \(n\) 的正整数序列分为 \(m\) 段，问你这 \(m\) 段最小的方差 \(v\) 为多少。输出 \(v\times m^2\) 。

\(1\leq n\leq 3000\)

Solution

容易发现答案就是：

\[m^2\frac{\sum\limits_{i=1}^m(a_i-\overline{a})^2}{m}=m\sum\limits_{i=1}^m\left(a_i-\frac{\sum\limits_{i=1}^m a_i}{m}\right)^2\]

记 \(s=\sum\limits_{i=1}^m a_i\) ，

\[m\sum_{i=1}^m a_i^2-2s\sum_{i=1}^m a_i+s^2\]

这玩意就可以斜率优化了。

upd

做的时候制杖了...发现最后的式子还可以化成 \[m\sum_{i=1}^m a_i^2-s^2\]

Code

//It is made by Awson on 2018.3.19

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 3000;

void read(LL &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

LL n, m, sum[N+5], s;

LL f[N+5][N+5];

int q[N+5], head, tail;

LL deltax(int p, int q) {return m*2*(sum[q]-sum[p]); }

LL deltay(int p, int q) {return m*sum[q]*sum[q]-m*sum[p]*sum[p]+s*2*sum[q]-s*2*sum[p]; }

void work() {

    read(n), read(m); for (int i = 1; i <= n; i++) read(sum[i]), sum[i] += sum[i-1];

    s = sum[n]; memset(f, 127/3, sizeof(f));

    f[0][0] = 0;

    for (int i = 1; i <= m; i++) {

    head = tail = 0; q[tail++] = 0;

    for (int j = 1; j <= n; j++) {

        while (tail-head > 1 &&

           f[i-1][q[head+1]]-f[i-1][q[head]]+deltay(q[head], q[head+1]) <=

           sum[j]*deltax(q[head], q[head+1])) ++head;

        f[i][j] = f[i-1][q[head]]+m*(sum[j]-sum[q[head]])*(sum[j]-sum[q[head]])-(sum[j]-sum[q[head]])*2*s;

        while (tail-head > 1 &&

           (f[i-1][q[tail-1]]-f[i-1][q[tail-2]]+deltay(q[tail-2], q[tail-1]))*deltax(q[tail-1], j) >=

           (f[i-1][j]-f[i-1][q[tail-1]]+deltay(q[tail-1], j))*deltax(q[tail-2], q[tail-1])) --tail;

        q[tail++] = j;

    }

    }

    writeln(f[m][n]+s*s);

}

int main() {

    work(); return 0;

}

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