BZOJ 1010: [HNOI2008]玩具装箱toy

1010

思路：

斜率优化dp

s[i]表示1-i的前缀和

斜率不等式为:

对于 i < j < k

（dp[j] - dp[k] + (j + s[j])^2 - (k + s[k])^2) / ((j + s[j]) - (k + s[k]))  <= 2*(i + s[i] - l -1)

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 5e4 + 5;
int c[N], l;
LL sum[N], dp[N];
LL get_down(int j, int k) {
    return (j + sum[j] - (k + sum[k]));
}
LL get_up(int j, int k) {
    return dp[j] - dp[k] + (j + sum[j])*(j + sum[j]) - (k + sum[k])*(k + sum[k]);
}
int main() {
    int n;
    scanf("%d %d", &n, &l);
    for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
    for (int i = 1; i <= n; i++) {
        sum[i] = sum[i-1] + c[i];
    }
    deque<int>q;
    q.push_front(0);
    for (int i = 1; i <= n; i++) {
        while(q.size() >= 2) {
            int t = q.front();
            q.pop_front();
            int tt = q.front();
            if(get_up(tt, t) <= 2*(i + sum[i] - l - 1)*get_down(tt, t));
            else {
                q.push_front(t);
                break;
            }
        }
        int t = q.front();
        dp[i] = dp[t] + (i - t - 1 + sum[i] - sum[t] - l) * (i - t - 1 + sum[i] - sum[t] - l);
        while(q.size() >= 2) {
            int t = q.back();
            q.pop_back();
            int tt = q.back();
            if(get_up(i, t) * get_down(t, tt) <= get_up(t, tt) * get_down(i, t)) ;
            else {
                q.push_back(t);
                break;
            }
        }
        q.push_back(i);
    }
    printf("%lld
", dp[n]);
    return 0;
}