BZOJ 3156: 防御准备(斜率优化dp)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
ll dp[1000010];
ll n;
ll a[1000010];
ll q[1000010];
double slope(ll j,ll k){
    return ((2.0*dp[j]+j+j*j)-(2.0*dp[k]+k+k*k))/(2.0*(j-k));
}
int main(){
    ios::sync_with_stdio(false);
    cin>>n; //一定要单组读入.. 怪异 
        for(int i=1;i<=n;i++)
            cin>>a[i];
        int l,r;
        l=r=1;
        for(int i=1;i<=n;i++){
            while(l<r&&slope(q[l],q[l+1])<i) l++;
            dp[i]=dp[q[l]]+(i-q[l]-1)*(i-q[l])/2+a[i]; //等差求和公式 
            while(l<r&&slope(q[r-1],q[r])>slope(q[r],i)) r--;
            q[++r]=i;
        }
        cout<<dp[n]<<endl;
    return 0;
}