BZOJ3437: 小P的牧场

设f[i]表示控制了1~i，第i个牧场建立了控制站，最小花费
推一下柿子
设第i个牧场的放养量是ai
f[i]=f[j]+∑j<k<i(i−k)ak
f[i]=f[j]+i∗∑j<k<iak−∑j<k<ik∗ak
设Si=∑ij=1aj，Vi=∑ij=1aj∗j
那么有
f[i]=f[j]+i∗(Si−1−Sj)−(Vi−1−Vj)
f[i]=i∗Si−1−Vi−1+f[j]−i∗Sj+Vj
若j<k，对于f[i]来说j更优时有
f[j]−i∗Sj+Vj<f[k]−i∗Sk+Vk
因为Sj<Sk于是
(f[j]+Vj)−(f[k]+Vk)(Sj−Sk)>i
这就是一个斜率优化了
所以维护一个斜率递增的凸包即可

读入格式都看错了竟然过了样例..

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long longusing namespace std;inline void read(ll &x){    char c;    while(!((c=getchar())>='0'&&c<='9'));    x=c-'0';    while((c=getchar())>='0'&&c<='9') (x*=10)+=c-'0';}const int maxn = 1000010;struct node{    ll x,y;    int i;    node(){}    node(const ll _x,const ll _y,const int _i){x=_x;y=_y;i=_i;}}q[maxn]; int head,tail;inline int multi(node p1,node p2,node p3){    p1.x-=p3.x; p1.y-=p3.y;    p2.x-=p3.x; p2.y-=p3.y;    return p1.x*p2.y-p1.y*p2.x;}inline bool cmp(node x,node y,ll k){return (x.y-y.y) < k*(x.x-y.x); }//y.x-x.x==0?int n;ll a[maxn],c[maxn];ll f[maxn],sck[maxn],sckk[maxn];int main(){    scanf("%d",&n); head=1,tail=1; q[1]=node(0,0,0);    for(int i=1;i<=n;i++) read(a[i]);    for(int i=1;i<=n;i++) read(c[i]);    for(ll i=1;i<=n;i++)    {        sck[i]=sck[i-1]+c[i]; sckk[i]=sckk[i-1]+c[i]*i;        while(head+1<=tail&&!cmp(q[head],q[head+1],i)) head++;        int j=q[head].i;        f[i]=a[i]+f[j]+i*(sck[i-1]-sck[j])-(sckk[i-1]-sckk[j]);        node tmp=node(sck[i],sckk[i]+f[i],i);        while(head+1<=tail&&multi(q[tail],tmp,q[tail-1])<=0) tail--;        q[++tail]=tmp;    }    printf("%lld\n",f[n]);    return 0;}