bzoj 3437 小K的农场 斜率优化

   没想到，最终还是太弱了。自己写的没4毫秒就WA了。翻了一下别人的。发现思路完全不一样。所以我想把自己的思路说出来，和大家分享一下。也希望帮我指正。

首先对于该题。我设 S[i] 的值为所有在 i 前面的点到 i 所需的费用和。那么我写出来的状态转移方程便是 f[i] = min(f[i], f[j] + tot[i] - tot[j+1] * (j - i))+a[i] (j <= i )   

最终得出的不等式就是(设j比k优,且k < j) f[j] + (t[j+1]*i) - f[k] - (t[k+1]*k) <= (t[j+1]-t[k+1])*i

然而WA了。那么正解就是差分维护前缀和了。（这不是重点，重点是我错哪了？）

#include<cstdio>
#include<iostream>
#define rep(i,j,k) for(register int i = j; i <= k; i++)
#define ll long long
#define maxn 1002333
using namespace std;

inline int read() {
    int s = 0, t = 1; char c = getchar();
    while( !isdigit(c) ) { if( c == '-' ) t = -1; c = getchar(); }
    while( isdigit(c) ) s = s * 10 + c - 48, c = getchar();
    return s * t;
}

ll f[maxn], y[maxn], ts[maxn], s[maxn];
int q[maxn];

int main() {
    //freopen("in.txt","r",stdin);
    int n = read(), x;
    rep(i,1,n) f[i] = read(); 
    ts[0] = s[0] = 0; rep(i,1,n) { s[i] = (x = read()) + s[i-1]; ts[i] = y[i] = 1ll * i * x + y[i-1]; }
    //rep(i,1,n) cout<<ts[i]<<" "; cout<<endl;
    //rep(i,1,n) 
    int l = 0, r = 0; q[0] = 0;
    rep(i,1,n){
        //cout<<"WORK "<<i<<endl;
        //rep(j,l,r) cout<<q[j]<<" "; cout<<endl;
        while( l < r && y[q[l+1]]-y[q[l]] <= (s[q[l+1]]-s[q[l]]) * i ) l++;
        //rep(j,l,r) cout<<q[j]<<" "; cout<<endl;
        f[i] += f[q[l]] + (s[i]-s[q[l]])*i - ts[i] + ts[q[l]]; y[i] += f[i];
        //cout<<f[i]<<" "<<y[i]<<endl;
        while( l < r && (y[q[r]]-y[q[r-1]])*(s[i]-s[q[r]]) >= (y[i]-y[q[r]])*(s[q[r]]-s[q[r-1]]) ) r--;
        q[++r] = i;
        //rep(j,l,r) cout<<q[j]<<" "; cout<<endl;
    }
    cout<<f[n]<<endl;
    return 0;
}