BZOJ1096仓库建设 and BZOJ3437小P的牧场 and BZOJ3156防御准备

这两道都是斜率优化的题目，完全一样的思路，双倍经验当然要收（就是读入坑了我好几次）。
设$f[i]$为在$i$位置建设了仓库的最小代价，则有：
$f[i]=\min(f[j]+x[i]*\sum_{l=j+1}^ip[l]-\sum_{l=j+1}^i(x[l]\times p[l]))+c[i] (0\leq j<i)$
前缀和优化一下，将原式转为：
$f[i]=\min(f[j]+x[i]\times (s2[i]-s2[j])-(s1[i]-s1[j]))+c[i](0\leq j<i)$
最后得：
$\frac{(f[i]+s1[j])-(f[k]+s1[k])}{s2[j]-s2[k]}<x[i]$

上代码：

#include<bits/stdc++.h>
#define il inline
#define int long long
using namespace std;
const int N=1000005;
#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
il int read(){
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=(x+(x<<2)<<1)+c-48;
    return x*f;
}
int n,h,t,x[N],p[N],c[N],f[N],s1[N],s2[N],q[N];
il int Y(int i){return f[i]+s1[i];}
il int X(int i){return s2[i];}
il double slope(int j,int k){return 1.0*(Y(j)-Y(k))/(X(j)-X(k));}
signed main(){
    n=read();
    for(int i=1;i<=n;++i) 
    x[i]=read(),p[i]=read(),c[i]=read(),s2[i]=s2[i-1]+p[i],s1[i]=s1[i-1]+x[i]*p[i];
    for(int i=1;i<=n;++i){
        while(h<t&&slope(q[h+1],q[h])<x[i]) ++h;
        int j=q[h];
        f[i]=f[j]+x[i]*(s2[i]-s2[j])-s1[i]+s1[j]+c[i];
        while(h<t&&slope(q[t],q[t-1])>slope(i,q[t])) --t;
        q[++t]=i;
    }
    return !printf("%lld",f[n]);
}

而小P的牧场就可以使$x[i]=i$，然后注意一下输入（其实没什么坑），就一样了。

#include<bits/stdc++.h>
#define il inline
#define int long long
using namespace std;
const int N=1000005;
#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
il int read(){
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=(x+(x<<2)<<1)+c-48;
    return x*f;
}
int n,h,t,p[N],c[N],f[N],s1[N],s2[N],q[N];
il int Y(int i){return f[i]+s1[i];}
il int X(int i){return s2[i];}
il double slope(int j,int k){return 1.0*(Y(j)-Y(k))/(X(j)-X(k));}
signed main(){
    n=read();
    for(int i=1;i<=n;++i) c[i]=read();
    for(int i=1;i<=n;++i) p[i]=read(),s2[i]=s2[i-1]+p[i],s1[i]=s1[i-1]+i*p[i];
    for(int i=1;i<=n;++i){
        while(h<t&&slope(q[h+1],q[h])<i) ++h;
        int j=q[h];
        f[i]=f[j]+i*(s2[i]-s2[j])-s1[i]+s1[j]+c[i];
        while(h<t&&slope(q[t],q[t-1])>slope(i,q[t])) --t;
        q[++t]=i;
    }
    return !printf("%lld",f[n]);
}

---

刚又发现防御准备这道题又是一样的，博客前面的就不改了，其实是三倍经验啊，只要把$x[i]$改为$i$,把$p[i]$改为就好。

#include<bits/stdc++.h>
#define il inline
#define int long long
using namespace std;
const int N=1000005;
#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
char buf[1<<21],*p1=buf,*p2=buf;
il int read(){
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=(x+(x<<2)<<1)+c-48;
    return x*f;
}
int n,h,t,c[N],f[N],s1[N],s2[N],q[N];
il int Y(int i){return f[i]+s1[i];}
il int X(int i){return s2[i];}
il double slope(int j,int k){return 1.0*(Y(j)-Y(k))/(X(j)-X(k));}
signed main(){
    n=read();
    for(int i=1;i<=n;++i) 
    c[i]=read(),s2[i]=s2[i-1]+1,s1[i]=s1[i-1]+i*1;
    for(int i=1;i<=n;++i){
        while(h<t&&slope(q[h+1],q[h])<i) ++h;
        int j=q[h];
        f[i]=f[j]+i*(s2[i]-s2[j])-s1[i]+s1[j]+c[i];
        while(h<t&&slope(q[t],q[t-1])>slope(i,q[t])) --t;
        q[++t]=i;
    }
    return !printf("%lld",f[n]);
}