P3994 高速公路

题目链接

题意分析

这是一道树上斜率优化题

首先

\[dp[i]=min\{dp[j]+(dis[i]-dis[j])* p[i]+q[i]\}(j∈Pre_i)
\]

那么就是

\[p[i]=\frac{dp[i]-dp[j]}{dis[i]-dis[j]}
\]

我们根据\(p[i]\)递增可知

我们需要使用单调队列维护下凸包

但是由于是树 所以我们不可以进行一般的单调队列维护

所以这里我们只好进行暴力二分出最优的位置了

毕竟单调队列实现起来不是真正的删除 而是头尾指针的移动 以及唯一的取代

然后我们会回溯时恢复指针以及取代值就可以了

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<queue>
#include<map>
#include<stack>
#include<list>
#include<set>
#include<deque>
#include<vector>
#include<ctime>
#define ll long long
#define inf 0x7fffffff
#define N 1008611
#define IL inline
#define M 1008611
#define D double
#define ull unsigned long long
#define R register
using namespace std;
template<typename T>IL void read(T &_)
{
    T __=0,___=1;char ____=getchar();
    while(!isdigit(____)) {if(____=='-') ___=0;____=getchar();}
    while(isdigit(____)) {__=(__<<1)+(__<<3)+____-'0';____=getchar();}
    _=___ ? __:-__;
}
/*-------------OI使我快乐-------------*/
ll n,tot,ans,head,tail;
ll to[N],nex[N],lat[N],w[N];
ll fa[N],p[N],q[N];
ll dis[N],dp[N],que[N];
IL void add(ll x,ll y,ll z)
{to[++tot]=y;nex[tot]=lat[x];lat[x]=tot;w[tot]=z;}
IL D cdy(int x,int y){return dis[y]-dis[x];}
IL D wzy(int x,int y){return dp[y]-dp[x];}
IL void dfs(ll now,ll fat)
{
    ll now_cdy=head,now_wzy=tail;
    ll le=now_cdy,ri=now_wzy-2,ans=-1;
    while(le<=ri)
    {
        ll mid=(le+ri)>>1;
        if(wzy(que[mid],que[mid+1])>=p[now]*cdy(que[mid],que[mid+1])) ri=mid-1,ans=mid;
        else le=mid+1;
    }
    if(ans!=-1) head=ans;
    else head=tail-1;
    dp[now]=dp[que[head]]+(dis[now]-dis[que[head]])*p[now]+q[now];
    le=head;ri=tail-2;ans=-1;
    while(le<=ri)
    {
        ll mid=(le+ri)>>1;
        if(wzy(que[mid],que[mid+1])*cdy(que[mid+1],now)<=wzy(que[mid+1],now)*cdy(que[mid],que[mid+1])) ans=mid,le=mid+1;
        else ri=mid-1;
    }
    if(ans!=-1) tail=ans+2;
    else tail=head+1;
    ll fro=que[tail];que[tail++]=now;
    for(R ll i=lat[now];i;i=nex[i])
    {
        ll v=to[i];
        if(v==fat) continue;
        dis[v]=dis[now]+w[i];
        dfs(v,now);
    }
    head=now_cdy;que[tail-1]=fro;tail=now_wzy;
}
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
    read(n);
    for(R ll i=2,x,y;i<=n;++i)
    {
        read(x);read(y);
        add(x,i,y);
        read(p[i]);read(q[i]);
    }
    for(R ll i=lat[1];i;i=nex[i])
    {
        ll v=to[i];
        dis[v]=w[i];
        que[head=0]=1;tail=1;
        dfs(v,1);
    }
    for(R ll i=2;i<=n;++i)
    printf("%lld\n",dp[i]);
//	fclose(stdin);
//	fclose(stdout);
    return 0;
}

HEOI 2019 RP++