BZOJ 1911: [Apio2010]特别行动队 [斜率优化DP]

1911: [Apio2010]特别行动队

Time Limit: 4 Sec  Memory Limit: 64 MB
Submit: 4142  Solved: 1964
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Description

Input

Output

Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

9

HINT

f[i]=max{f[j]+...}

随便一化就好了

(a*(s[k]*s[k]-s[j]*s[j])+f[k]-f[j]+b*(s[j]-s[k])) / (2*a*(s[k]-s[j]))

最后是s[i]>=slope(j,k)时k优

s[]是单调的，用单调队列维护这个下凸壳

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=1e6+5,INF=1e9;
typedef long long ll;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();}
    return x*f;
}
int n,a,b,c;
ll s[N],f[N];
inline double slope(int j,int k){
    return  (double)(a*(s[k]*s[k]-s[j]*s[j])+f[k]-f[j]+b*(s[j]-s[k]))/(double)(2*a*(s[k]-s[j]));
}
int q[N],head,tail;
void dp(){
    head=tail=1;
    for(int i=1;i<=n;i++){
        while(head<tail&&slope(q[head],q[head+1])<=s[i]) head++;
        int j=q[head];
        f[i]=f[j]+a*(s[i]-s[j])*(s[i]-s[j])+b*(s[i]-s[j])+c;//printf("f %lld %d\n",f[i],j);
        while(head<tail&&slope(q[tail-1],q[tail])>slope(q[tail],i)) tail--;
        q[++tail]=i;
    }
    printf("%lld",f[n]);
}
int main(){
    //freopen("in.txt","r",stdin);
    n=read();a=read();b=read();c=read();
    for(int i=1;i<=n;i++) s[i]=s[i-1]+read();
    dp();
}