bzoj 1010: [HNOI2008]玩具装箱toy

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1010

学习斜率dp的话请翻看我的 dp斜率优化小计

递推公式：f[i]=min(f[j]+(i-j-1+sum[i]-sum[j]-L)^2) , j<i

令g[i]=sum[i]+i, c=L+1

有f[i]=min(f[j]+(g[i]-g[j]-L)^2), j<i

展开，化成f[i]=min(-a[i]*x[j]+y[j])+w[i]形式有：

a[i]=2*g[i]

x[j]=g[j]

y[j]=f[j]+g[j]^2+2*g[j]*c

可以看出，a[i]与(x[j], y[j])均为增序列

那么用单调队列维护凸包就好了

/*
 * Problem:  BZOJ 1010
 * Author:  SHJWUDP
 * Created Time:  2015/4/28 星期二 14:09:04
 * File Name: 233.cpp
 * State: Accepted
 * Memo: 斜率优化dp
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int64;
typedef unsigned long long int65;

const int MaxA=5e4+7;

int N, L;
int C[MaxA];
int64 sum[MaxA], g[MaxA];
int64 f[MaxA];
int deque[MaxA];
inline int64 getDX(int i, int j) {
    return g[i]-g[j];
}
inline int64 getDY(int i, int j) {
    return (f[i]+g[i]*g[i]+2*(L+1)*g[i])-(f[j]+g[j]*g[j]+2*(L+1)*g[j]);
}
inline int64 cal(int i, int j) {
    return f[j]+(g[i]-g[j]-(L+1))*(g[i]-g[j]-(L+1));
}
void solve() {
    f[0]=0;
    int front=0, tail=0;
    deque[tail++]=0;
    for(int i=1; i<=N; i++) {
        //如果单调队列中第后一个比前一个更优，那么舍弃前一个
        while(front+1<tail && cal(i, deque[front+1])<=cal(i, deque[front])) front++;
        f[i]=cal(i, deque[front]);
        //维护下凸包
        while(front+1<tail && 
                getDY(i, deque[tail-1])*getDX(deque[tail-1], deque[tail-2])
                <= getDY(deque[tail-1], deque[tail-2])*getDX(i, deque[tail-1])) tail--;
        deque[tail++]=i;
    }
    printf("%lld
", f[N]);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif
    while(~scanf("%d%d", &N, &L)) {
        sum[0]=0;
        for(int i=1; i<=N; i++) {
            scanf("%d", &C[i]);
            sum[i]=sum[i-1]+C[i];
            g[i]=sum[i]+i;
        }
        solve();
    }
    return 0;
}