特别行动队[APIO2010]

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1911: [Apio2010]特别行动队

Time Limit: 4 Sec  Memory Limit: 64 MB
Submit: 5349  Solved: 2651
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Description

Input

Output

Sample Input

4
-1 10 -20
2 2 3 4

Sample Output

9

HINT

Source

 

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　　 斜率优化dp

　　不过需要注意的是这道题斜率是负的，并且维护的是个上凸包。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std ;
const int N = 1e6 + 10 ;

inline LL read() {
    LL k = 0, f = 1 ; char c = getchar() ;
    for( ; !isdigit(c) ; c = getchar())
      if(c == '-') f = -1 ;
    for( ; isdigit(c) ; c = getchar())
      k = k*10 + c-'0' ;
    return k*f ;
} 
int n ; LL a, b, c ; LL pp[N] ;
int q[N] ; LL f[N] ;

inline LL sqr(LL x) { return x*x ; }
inline double Y(int i) { return f[i]+a*sqr(pp[i])-b*pp[i] ; }
inline double X(int i) { return pp[i] ; }
inline double Rate(int i,int j) { return (Y(j)-Y(i))/(X(j)-X(i)) ; }

int main() {
    n = read() ; a = read(), b = read(), c = read() ;
    for(int i=1;i<=n;i++) pp[i] = pp[i-1]+read() ;
    int head = 1, tail = 1 ; q[1] = 0 ;
    for(int i=1;i<=n;i++) {
        while(head < tail && Rate(q[head],q[head+1]) > (double)2*a*pp[i]) head++ ;
        int j = q[head] ; f[i] = f[j]+a*sqr(pp[i]-pp[j])+b*(pp[i]-pp[j])+c ;
        while(head < tail && Rate(q[tail-1],q[tail]) < Rate(q[tail],i)) tail-- ;
        q[++tail] = i ;
    }
    printf("%lld",f[n]) ;
    return 0 ;
}