[bzoj1010][HNOI2008]玩具装箱

Description

有\(n\)个物品,每个物品长度为\(c_i\),现在要把这\(n\)个物品划分成若干组,每组中的物品编号是连续的,规定每组的长度\(x=j-i+\sum_{k=i}^{j}c_k\),费用为\((x-l)^2\),求最小费用.

Input

第一行输入两个整数\(n,l\),接下来\(n\)行输入\(c_i\).

Output

一行表示最小费用.

Sample Input

5 4
3
4
2
1
4

Sample Output

1

HINT

\(1\;\leq\;n\;\leq\;50000,1\;\leq\;l,c_i\;\leq\;10^7\)

Solution

\(f[i]\)表示将前i个物品分组所需最小费用.

\(f[i]=min\{f[j]+(i-j-1+sum[i]-sum[j]-l)^2\}(j<i)\)

\(O(n^2)\)会\(T\),考虑斜率优化.

当\(k>j\)且\(f[i]_k<f[i]_j\)时,
\(f[i]_j=f[j]+(i-j-1+sum[i]-sum[j]-l)^2\)
\(f[i]_k=f[k]+(i-k-1+sum[i]-sum[k]-l)^2\)
\(\\\)
尽量将\(i,j\)分离,设\(g(j)=j+sum[j],h(i)=i+sum[i]-1-l\),得
\(f[i]_j=f[j]+(h(i)-g(j))^2\)
\(f[i]_k=f[k]+(h(i)-g(k))^2\)
\(\\\)
\(f[i]_k<f[i]_j\)的前提是
\(f[k]-f[j]+(h(i)-g(k))^2-(h(i)-g(j))^2<0\)
整理得\(\frac{(f[k]+g(k)^2)-(f[j]+g(j)^2)}{g(k)-g(i)}<2\;\times\;h(i)\)
\(\\\)
设\(T(j_1,j_2)=\frac{(f[j_2]+g(j_2)^2)-(f[j_1]+g(j_1)^2)}{g(j_2)-g(j_1)}\)
则\(T(j_1,j_2)<T(j_2,j_3)<...\)
(若存在\(T(j_{n-1},j_n)>T(j_n,j_{n+1}\)),因为\(h(i)\)单调递增,所以\(j_{n+1}\)一定比\(j_n\)优,即\(j_n\)可以删去)
所以每次取元素时,将满足\(T(j_n,j_{n+1})<2\;\times\;h(i)\)的\(j_n\)出队(因为\(j_{n+1}\)比\(j_n\)优),然后取队首为\(j\).

#include<set> 
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 50001
using namespace std;
typedef long long ll;
ll f[N],s[N],q[N],h,t,l,n;
inline ll sqr(ll k){
    return k*k;
}
inline ll x(ll k){
    return k+s[k];
}
inline ll y(ll k){
    return f[k]+sqr(x(k));
}
inline double cmp(ll p,ll q){
    return (double)(y(q)-y(p))/(double)(x(q)-x(p));
}
inline ll g(ll k){
    return k+s[k]-l;
}
inline void init(){
    scanf("%lld%lld",&n,&l);
    for(ll i=1;i<=n;i++){
        scanf("%lld",&s[i]);
        s[i]+=s[i-1];
    }
    for(ll i=1,k;i<=n;i++){
        k=g(i)<<1;
        while(h<t&&cmp(q[h],q[h+1])<k) h++;
        f[i]=f[q[h]]+sqr(x(q[h])-g(i)+1);
        while(h<t&&cmp(q[t],i)<cmp(q[t-1],q[t]))
            t--;
        q[++t]=i;
    }
    printf("%lld\n",f[n]);
}
int main(){
    freopen("toy.in","r",stdin);
    freopen("toy.out","w",stdout);
    init();
    fclose(stdin);
    fclose(stdout);
    return 0;
}