APIO 2010 特别行动队 | 斜率优化DP

luogu 3628

si表示序列的前缀和
f(i)表示将序列的前i个划分若干段的最大价值
f(i)= max{f(j)+a∗(si−sj)2+b∗(si−sj)+c},1≤j＜i
    = max{−2a*sj*si+f(j)+a*sj*sj−b*sj}+a*si*si+b*si+c,1≤j＜i

 1 #include <cstdio>
 2 #include <string>
 3
 4 typedef long long ll;
 5
 6 ll a, b, c;
 7
 8 ll read() {
 9     ll x = 0, f = 1;
10     char c = getchar();
11     while (!isdigit(c)) {
12         if (c == ‘-‘) f = -1;
13         c = getchar();
14     }
15     while (isdigit(c)) {
16         x = (x << 3) + (x << 1) + (c ^ 48);
17         c = getchar();
18     }
19     return x * f;
20 }
21
22 ll s[1000005], f[1000005]; int Q[1000005];
23
24 ll K(int j) {
25     return -2 * a * s[j];
26 }
27
28 ll B(int j) {
29     return f[j] + a * s[j] * s[j] - b * s[j];
30 }
31
32 ll Y(int i, int j) {
33     return K(j) * s[i] + B(j);
34 }
35
36 bool cover(int i, int j, int k) {
37     ll y1 = (K(i) - K(k)) * (B(j) - B(i));
38     ll y2 = (K(i) - K(j)) * (B(k) - B(i));
39     return y1 <= y2;
40 }
41
42 int main() {
43     int n = read(); a = read(), b = read(), c = read();
44     for (int i = 1; i <= n; ++ i) {
45         ll x = read(); s[i] = s[i - 1] + x;
46     }
47     int l = 0, r = 0;
48     for (int i = 1; i <= n; ++ i) {
49         while (l < r && Y(i, Q[l]) <= Y(i, Q[l + 1])) ++ l;
50         f[i] = Y(i, Q[l]) + a * s[i] * s[i] + b * s[i] + c;
51         while (l < r && cover(i, Q[r], Q[r - 1])) -- r;
52         Q[++r] = i;
53     }
54     printf("%lld\n", f[n]);
55 }

原文地址：https://www.cnblogs.com/milky-w/p/8541609.html