[APIO 2010] 特别行动队

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=1911

[算法]

设前i个士兵"修正"后的最大战斗力为fi

令sumi表示x的前缀和

显然 ， 有状态转移方程 : fi = max{ fj + a * (sumi - sumj) ^ 2 + b * (sumi - sumj) + c }

对该式进行化简 ， 得 :

fi = max{ fj + asumi ^ 2 + asumj ^ 2 - 2asumisumj + bsumi - bsumj + c}

令Yj = fj + asumj ^ 2 , Xj = sumj

则 : fi = max{Yj - Xj(2sumi + b) + aumi ^ 2 + bsumi + c}

那么Yj = Xj + (2asumi + b) + fi - asumi ^ 2 - bsumi - c

显然我们要做的是最大化截距

2asumi + b单调递减 ， Xi单调递增 ， 维护一个上凸壳即可

时间复杂度 : O(N)

[代码]

#include<bits/stdc++.h>
using namespace std;
const int N = 1000010;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

int n , l , r;
ll a , b , c;
int q[N];
ll f[N] , sum[N] , X[N] , Y[N];

template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}

int main()
{

        read(n);
        read(a); read(b); read(c);
        for (int i = 1; i <= n; i++)
        {
                int x;
                read(x);
                sum[i] = sum[i - 1] + x;
                X[i] = sum[i];
        }
        f[q[l = r = 1] = 0] = 0;
        for (int i = 1; i <= n; i++)
        {
                while (l < r && Y[q[l + 1]] - Y[q[l]] >= (2 * a * sum[i] + b) * (X[q[l + 1]] - X[q[l]])) ++l;
                f[i] = Y[q[l]] - X[q[l]] * (2 * a * sum[i] + b) + a * sum[i] * sum[i] + b * sum[i] + c;
                Y[i] = f[i] + a * sum[i] * sum[i];
                while (l < r && (Y[i] - Y[q[r]]) * (X[q[r]] - X[q[r - 1]]) >= (Y[q[r]] - Y[q[r - 1]]) * (X[i] - X[q[r]])) --r;
                q[++r] = i;
        }
        printf("%lld\n" , f[n]);

        return 0;

}

原文地址：https://www.cnblogs.com/evenbao/p/10354200.html