1911: [Apio2010]特别行动队

Time Limit: 4 Sec Memory Limit: 64 MB
Submit: 5706 Solved: 2876
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Description

Input

Output

Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

HINT

似乎逐渐掌握了斜率优化的规律，，，

f[i]=max{f[j]+a*(sum[i]-sum[j])²+b*(sum[i]-sum[j]+c}

斜率优化，设k<j<i，且j比k更优

[f[i]-f[k]+a(s[j]^{2}-s[k]^{2})+b(s[k]-s[j])>2a imes s[i] imes (s[j]-s[k])]

化为：（注意a为负数！！！要变号）

[frac{f[i]-f[k]+a(s[j]^{2}-s[k]^{2})+b(s[k]-s[j])}{2a(s[j]-s[k])}<s[i]]

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 
 5 #define LL long long
 6 #define sqr(x) ((x)*(x))
 7 const int MAXN=1000005;
 8 
 9 int n,a,b,c;
10 int l,r,q[MAXN];
11 LL f[MAXN],s[MAXN];
12 
13 LL cal(LL x)
14 {
15     return a*sqr(x)+b*x+c;
16 }
17 
18 double slope(int k,int j)
19 {
20     return (double)(f[j]-f[k]+a*(sqr(s[j])-sqr(s[k]))+b*(s[k]-s[j]))/(2*a*(s[j]-s[k]));
21 }
22 
23 int main()
24 {
25     scanf("%d",&n);
26     scanf("%d%d%d",&a,&b,&c);
27     for(int i=1;i<=n;i++)
28     {
29         int x;
30         scanf("%d",&x);
31         s[i]=s[i-1]+x;
32     }
33     for(int i=1;i<=n;i++)
34     {
35         while(l<r&&slope(q[l],q[l+1])<s[i]) l++;
36         int t=q[l];
37         f[i]=f[t]+cal(s[i]-s[t]);
38         while(l<r&&slope(q[r],i)<slope(q[r-1],q[r])) r--;
39         q[++r]=i;
40     }
41     printf("%lld",f[n]);
42     return 0;
43 }