Interpretations P2605 [[ZJOI2010]] base station site (From luoguBlog)

Segment tree optimization dp

The array F [i] [j] denotes the i-th first villages, the j-th base stations built at a minimum cost i

The intersection line regressed and Wang Hesong method we can obtain the equation

f[i][j]=min(f[k][j−1]+cost(k,i))

cost (k, i) represents the cost of compensation required villages between the first base station coverage is not i ~ k villages, cost calculation complexity is O (n)

Pretreatment using binary search range of requirements for each location bef [i], beh [i]

After the minimum is to use the tree line maintenance f [] + cost (), the query interval interval Update

When beh [x] = i, if i is not built, the added cost (there may be a lot of x, star or forward vector storage)

Code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#define ls(k) k<<1
#define rs(k) k<<1|1
using namespace std;
const int N=20010,K=110;
int dis[N],s[N],w[N],c[N],f[N];
int n,m,bef[N],beh[N];
int tot=0,to[N<<1],head[N<<1],nxt[N<<1];
int mn[N<<2],lz[N<<2];
void Add(int x,int y)
{
    to[++tot]=y;
    nxt[tot]=head[x];
    head[x]=tot;
}
void up(int k)
{
    mn[k]=min(mn[ls(k)],mn[rs(k)]);
}
void build(int k,int l,int r)
{
    lz[k]=0;
    if(l==r)
    {
        mn[k]=f[l];
        return ;
    }
    int mid=l+r>>1;
    build(ls(k),l,mid);
    build(rs(k),mid+1,r);
    up(k);
}
void down(int k)
{
    lz[ls(k)]+=lz[k];
    lz[rs(k)]+=lz[k];
    mn[ls(k)]+=lz[k];
    mn[rs(k)]+=lz[k];
    lz[k]=0;
}
int query(int k,int l,int r,int L,int R)
{
    if(L>R)return 0x3f3f3f3f;
    if(L<=l&&R>=r)return mn[k];
    int mid=l+r>>1;
    if(lz[k])down(k);
    int res=0x3f3f3f3f;
    if(L<=mid)res=min(res,query(ls(k),l,mid,L,R));
    if(mid<R)res=min(res,query(rs(k),mid+1,r,L,R));
    return res;
}
void change(int k,int l,int r,int L,int R,int vl)
{
    if(L>R)return ;
    if(L<=l&&R>=r)
    {
        lz[k]+=vl;
        mn[k]+=vl;
        return ;
    }
    if(lz[k])down(k);
    int mid=l+r>>1;
    if(L<=mid)change(ls(k),l,mid,L,R,vl);
    if(R>mid)change(rs(k),mid+1,r,L,R,vl);
    up(k);
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=2;i<=n;i++)scanf("%d",&dis[i]);
    for(int i=1;i<=n;i++)scanf("%d",&c[i]);
    for(int i=1;i<=n;i++)scanf("%d",&s[i]);
    for(int i=1;i<=n;i++)scanf("%d",&w[i]);
    n++,m++;
    dis[n]=w[n]=0x3f3f3f3f;
    for(int i=1;i<=n;i++)
    {
        bef[i]=lower_bound(dis+1,dis+n+1,dis[i]-s[i])-dis;
        beh[i]=lower_bound(dis+1,dis+n+1,dis[i]+s[i])-dis;
        if(dis[beh[i]]>dis[i]+s[i])beh[i]--;
        Add(beh[i],i);
    }
    int now=0;
    for(int j=1;j<=n;j++)
    {
        f[j]=now+c[j];
        for(int i=head[j];i;i=nxt[i])now+=w[to[i]];
    }
    int ans=f[n];
    for(int i=2;i<=m;i++)
    {
        build(1,1,n);
        for(int j=1;j<=n;j++)
        {
            f[j]=query(1,1,n,1,j-1)+c[j];
            for(int p=head[j];p;p=nxt[p])
                change(1,1,n,1,bef[to[p]]-1,w[to[p]]);
        }
        ans=min(ans,f[n]);
    }
    cout<<ans<<endl;
    return 0; 
}