【BZOJ 1010 [HNOI2008]玩具装箱toy】 斜率优化

bzoj 1010
说实话啊 一开始搞这题把自己心态搞崩了
dp状态好列
但是和高手的差距就是 我只是统计了前缀和
高手们都是统计了 sum[i] = arr[1] + arr[2] + … +arr[i] + i
这样我们是不是就避免了 i - j - 1？
很巧妙吧
为什么C要+1
因为我们默认是丛 j + 1 到 i
你从前缀和就能体现了
式子这样列
dp[i] = min(dp[i],dp[j] + (f[i]-f[j]-C)^2)
然后 设 k > i 并且 k 更优
推出 (Y[j]-Y[k])/(X[j] -X[k]) >= 2 * f[i]
其中 Y(x) = dp[x] + sum[x]*sum[x] + 2 * C * sum[x]
X(x) = sum[x]
维护一个下凸的就行了

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 50025;
ll sum[MAX_N],arr[MAX_N],dp[MAX_N],C;
int q[MAX_N],h = 0,t = 1;
ll getdp(int j,int k)
{
    return (dp[j]+sum[j]*sum[j]+2*C*sum[j])-(dp[k]+sum[k]*sum[k]+2*C*sum[k]);
}
ll getdown(int j,int k)
{
    return (sum[j] - sum[k]);
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    scanf("%d%lld",&n,&C);
    C++;
    for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]),sum[i] = sum[i-1]+arr[i]+1;
    for(int i = 1;i<=n;++i)
    {
        while(h<t&&getdp(q[h+1],q[h])<=2*sum[i]*getdown(q[h+1],q[h])) ++h;
        dp[i] = dp[q[h]] + (sum[i] - sum[q[h]] - C)*(sum[i] - sum[q[h]] - C);
        while(h<t&&getdp(i,q[t])*getdown(q[t],q[t-1])<=getdp(q[t],q[t-1])*getdown(i,q[t])) --t;
        q[++t] = i;
    }
    printf("%lld\n",dp[n]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}