jzoj 4475. 【GDOI2016模拟4.25】征途

Description

将n个数刚好分成m段，每段的贡献为此段数的和，求这m个数的最小方差*(m^2)
对于 100% 的数据，1≤n≤3000。

Solution

由于是连续一段的，一眼斜率优化DP。
设\(qz[i]\)表示\(a[1]\)~\(a[i]\)的和。
我们考虑一下将答案化简。设平均数为\(k\)。
\[ans=((x_1-k)^2+...+(x_m-k)^2)/m*m^2\]
\[ans=((x_1)^2+...+(x_m)^2-2*(x_1+...+x_m)*k+m*k^2)*m\]
\[ans=m*((x_1)^2+...+(x_m)^2)-2*qz[n]^2+qz[n]^2\]
\[ans=m*((x_1)^2+...+(x_m)^2)-qz[n]^2\]
因为要刚好分成\(m\)段，所以直接上凸优化。
我们二分一个数，最后答案便是\(f[n]-m*mid-qz[n]*qz[n]\)

Code

#include <cstdio>
#include <cstring>
#define N 3010
#define ll long long
#define mem(x, a) memset(x, a, sizeof x)
#define fo(x, a, b) for (int x = a; x <= b; x++)
using namespace std;
int n, m, qz[N], g[N], l, len;
ll f[N][2];

inline int read()
{
    int x = 0; char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x;
}

int sqr(int x) {return x * x;}

ll left(int x, int y) {return f[x][0] - f[y][0] + sqr(qz[x]) - sqr(qz[y]);}

ll right(int x, int y) {return 2 * (qz[x] - qz[y]);}

void check(int x)
{
    f[0][0] = 0, f[0][1] = 0;
    g[l = len = 1] = 0;
    fo(i, 1, n)
    {
        while (l < len && left(g[l + 1], g[l]) < right(g[l + 1], g[l]) * qz[i]) l++;
        f[i][0] = f[g[l]][0] + sqr(qz[i] - qz[g[l]]) + x;
        f[i][1] = f[g[l]][1] + 1;
        while (l < len && left(g[len], g[len - 1]) * right(i, g[len]) >= left(i, g[len]) * right(g[len], g[len - 1])) len--;
        g[++len] = i;
    }
}

int main()
{
    freopen("journey.in", "r", stdin);
    freopen("journey.out", "w", stdout);
    n = read(), m = read(); qz[0] = 0;
    fo(i, 1, n) qz[i] = qz[i - 1] + read();
    check(0);
    long long l = 0, r = 100000000000, mid;
    while (l <= r)
    {
        mid = l + r >> 1;
        check(mid);
        if (f[n][1] == m) break;
        if (f[n][1] < m) r = mid - 1;
        else l = mid + 1;
    }
    printf("%d\n", (f[n][0] - mid * m) * m - qz[n] * qz[n]);
    return 0;
}