[HNOI2008]玩具装箱toy（dp+斜率优化）

    斜率优化问题一般都是决策单调问题。对于这题能够证明单调决策。

令sum[i]=sigma(c [k] ) 1<=k<=i  ,  f[i]=sum[i]+i ,  c=L+1;

 首先我们能够写出转移方程  dp[i] = min( dp[j] + (f[i]-f[j]-c)^2 )  。令决策j1<j2。若决策j2更优有

 dp[j2]+(f[i]-f[j2]-c)^2<=dp[j1]+(f[i]-f[j1]-c)^2

能够得带 ((dp[j2]+f[j2]^2)-(dp[j1]+f[j1]^2)  )/(f[j2]-f[j1])<2*(f[i]-c)。

优于f[i]是递增的，所以对于t>i的点。决策j2总是比j1更优。那么j1实际上能够从决策集合中删除。后面的就能够用一个队列维护了。

<span style="font-size:14px;">#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <string>
#include <cctype>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int inf  = 0x3fffffff;
const int mmax =50010;
LL C[mmax];
LL L,c;
LL sum[mmax],f[mmax],dp[mmax];
LL sqr(LL x)
{return x*x;
}
double G(int x)
{return 1.0*f[x]*f[x]+dp[x];
}
double S(int x)
{return 2.0*f[x];
}
void calc(int i,int j)
{dp[i]=dp[j]+sqr(f[i]-f[j]-c);
}
int Q[mmax];
int main()
{int n;while(cin>>n>>L){c=L+1;sum[0]=0;f[0]=0;for(int i=1;i<=n;i++){scanf("%lld",&C[i]);sum[i]=sum[i-1]+C[i];f[i]=sum[i]+i;}int head=0,tail=-1;dp[0]=0;Q[++tail]=0;for(int i=1;i<=n;i++){while(head<tail){double tmp=1.0*(G(Q[head+1])-G(Q[head]))/(S(Q[head+1])-S(Q[head]));if(tmp<=f[i]-c)head++;elsebreak;}calc(i,Q[head]);while(head<tail){double tmp1=1.0*(G(Q[tail])-G(Q[tail-1]))/(S(Q[tail])-S(Q[tail-1]));double tmp2=1.0*(G(i)-G(Q[tail]))/(S(i)-S(Q[tail]));if(tmp1>=tmp2)tail--;elsebreak;}Q[++tail]=i;}printf("%lld\n",dp[n]);}return 0;
}
</span>