BZOJ 1911 [APIO2010] 特别行动队

题解

非常显然的$O(n^2)$ 的dp转移方程 :

$ F_i  = max( F_j +  a imes (S_i - S_j) ^2 + b imes (S_i - S_j ) + c) $ 数组S为前缀和

进行一波分离后变成了 $ F_j + a imes (S_j)^2 - b imes S_j = 2a imes S_i imes S_j - b imes S_i - c + F_i$。

左边的一串式子为纵坐标， $S_j$ 为横坐标， $2a imes  S_i  $当做斜率， 套上斜率优化板子就可以了

注意维护的是上凸壳

代码

#include<cstring>
#include<algorithm>
#include<cstdio>
#define rd read()
#define rep(i,a,b) for(int i = (a); i <= (b); ++i)
#define per(i,a,b) for(int i = (a); i >= (b); --i)
#define ll long long
using namespace std;
 
const int N = 1e6 + 1e5;
 
ll f[N], a, b, c, s[N], q[N], sum[N], n;
 
int read() {
    int X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if( c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}
 
double caly(int A) {
    return f[A] + a * sum[A] * sum[A] - b * sum[A];
}
 
double calk(int A, int B) {
    double ax = sum[A], bx = sum[B], ay = caly(A), by = caly(B);
    return (by - ay) / (bx - ax);
}
 
 
int main()
{
    n = rd;
    a = rd; b = rd; c = rd;
    rep(i, 1, n) s[i] = rd, sum[i] = sum[i - 1] + s[i];
    int l = 1, r = 1;
    rep(i, 1, n) {
        while(l < r && calk(q[l], q[l + 1]) >= 2 * a * sum[i]) l++;
        ll x = sum[i] - sum[q[l]];
        f[i] = f[q[l]] + a * x * x + b * x + c;
        while(l < r && calk(q[r - 1], q[r]) <= calk(q[r], i)) r--;
        q[++r] = i;
    }
    printf("%lld
", f[n]);
}