noi.ac #531 神树和物品

题目链接：戳我

决策单调性（证明？不会啊，自己打表看QAQ）

44pts \(O(n^2)\)代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MAXN 1010
using namespace std;
int n;
long long a[MAXN],b[MAXN],c[MAXN],f[MAXN];
inline long long pow3(long long x){return 1ll*x*x*x;}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("ce.in","r",stdin);
    #endif
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    for(int i=1;i<=n;i++) scanf("%lld",&b[i]);
    for(int i=1;i<=n;i++) scanf("%lld",&c[i]);
    for(int i=0;i<=n;i++) f[i]=0x3f3f3f3f3f3f3f3f;
    f[0]=0;
    for(int i=1;i<=n;i++)
        for(int j=0;j<i;j++)
            f[i]=min(f[i],f[j]+pow3(abs(a[i]-b[j+1]))+pow3(c[j+1]));
    printf("%lld\n",f[n]);
    return 0;
}

100pts代码：时间复杂度\(O(nlogn)\)（貌似跑的挺快的hhh）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MAXN 300010
using namespace std;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*f;
}
int n;
int a[MAXN],b[MAXN],c[MAXN];
struct Node{int pos,l,r;}q[MAXN<<1];
long long f[MAXN];
inline long long calc(int l,int r)
{
    long long cur_ans=1ll*(a[r]-b[l+1])*(a[r]-b[l+1])*(a[r]-b[l+1]);
    return f[l]+abs(cur_ans)+1ll*c[l+1]*c[l+1]*c[l+1];
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("ce.in","r",stdin);
    #endif
    n=read();
    for(int i=1;i<=n;i++) a[i]=read();
    for(int i=1;i<=n;i++) b[i]=read();
    for(int i=1;i<=n;i++) c[i]=read();
    int head=1,tail=1;
    q[head]=(Node){0,1,n};
    for(int i=1;i<=n;i++)
    {
        while(head<tail&&q[head].r<i) head++;
        f[i]=calc(q[head].pos,i);
        while(head<tail&&calc(i,q[tail].l)<=calc(q[tail].pos,q[tail].l)) tail--;
        int l=q[tail].l,r=q[tail].r,cur_ans=r+1;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(calc(i,mid)<=calc(q[tail].pos,mid)) cur_ans=mid,r=mid-1;
            else l=mid+1;
        }
        if(cur_ans!=q[tail].l) q[tail].r=cur_ans-1;
        else tail--;
        if(cur_ans<=n) q[++tail]=(Node){i,cur_ans,n};
    }
    printf("%lld\n",f[n]);
    return 0;
}

标签：ac,ch,noi,int,神树,long,MAXN,include,getchar
来源： https://www.cnblogs.com/fengxunling/p/11140454.html