bzoj3672: [Noi2014]购票（树形DP+斜率优化+可持久化凸包）

　　这题的加强版，多了一个$l_i$的限制，少了一个$p_i$的单调性，难了好多...

　　首先有方程$f(i)=min\{f(j)+(dep_i-dep_j)*p_i+q_i\}$

　　$\frac {f(j)-f(k)}{dep_j-dep_k}<p_i$

　　假如没有$l_i$的限制，实际上就是上面那题...

　　如果多了$l_i$的限制会有什么影响呢？

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#define ll long long
#define int long long
using namespace std;
const int maxn=500010, inf=1e9;
struct poi{int too, dis, pre;}e[maxn];
struct tjm{int bg, len;}tree[maxn<<2];
int n, m, x, z, tot, sum;
int v[maxn*20], last[maxn], p[maxn], l[maxn], memv[maxn][20], meml[maxn][20];
ll f[maxn], dep[maxn], deep[maxn], q[maxn];
inline void read(int &k)
{
int f=1; k=0; char c=getchar();
while(c<'0' || c>'9') c=='-'&&(f=-1), c=getchar();
while(c<='9' && c>='0') k=k*10+c-'0', c=getchar();
k*=f;
}
inline void add(int x, int y, int z){e[++tot]=(poi){y, z, last[x]}; last[x]=tot;}
inline double xl(int x, int y){return 1.0*(f[x]-f[y])/(deep[x]-deep[y]);}
inline int findans(int l, int r, ll limit)
{
if(l>=r) return r;
int L=l+1, R=r;
while(L<R)
{
int mid=(L+R)>>1;
if(limit-xl(v[mid], v[mid-1])>1e-9) L=mid+1;
else R=mid;
}
if(xl(v[L], v[L-1])>=limit) L--;
return L;
}
inline int find(int r, int limit)
{
int L=1, R=r;
while(L<R)
{
int mid=(L+R)>>1;
if(dep[mid]>=limit) R=mid;
else L=mid+1;
}
return L;
}
inline int findr(int l, int r, int pos)
{
if(l>=r) return r;
int L=l+1, R=r;
while(L<R)
{
int mid=(L+R+1)>>1;
if(xl(v[mid], v[mid-1])-xl(v[mid], pos)>1e-9) R=mid-1;
else L=mid;
}
if(xl(v[L], v[L-1])-xl(v[L], pos)>1e-9) L--;
return L;
}
void build(int x, int l, int r)
{
tree[x].bg=sum; sum+=r-l+1;
if(l==r) return;
int mid=(l+r)>>1;
build(x<<1, l, mid); build(x<<1|1, mid+1, r);
}
int query(int x, int l, int r, int cx, ll limit)
{
if(!tree[x].len || r<cx) return -1;
if(cx<=l) return findans(tree[x].bg, tree[x].bg+tree[x].len-1, limit);
int mid=(l+r)>>1, lt=query(x<<1, l, mid, cx, limit), rt=query(x<<1|1, mid+1, r, cx, limit);
if(lt!=-1 && rt!=-1) if(limit-xl(v[lt], v[rt])>1e-9) swap(lt, rt);
return lt==-1?rt:lt;
}
void update(int x, int l, int r, int cx, int pos, int d)
{
int R=findr(tree[x].bg, tree[x].bg+tree[x].len-1, pos);
memv[pos][d]=v[++R]; meml[pos][d]=tree[x].len; v[R]=pos; tree[x].len=R-tree[x].bg+1;
if(l==r) return;
int mid=(l+r)>>1;
if(cx<=mid) update(x<<1, l, mid, cx, pos, d+1);
else update(x<<1|1, mid+1, r, cx, pos, d+1);
}
void recovey(int x, int l, int r, int cx, int pos, int d)
{
v[tree[x].bg+tree[x].len-1]=memv[pos][d]; tree[x].len=meml[pos][d];
if(l==r) return;
int mid=(l+r)>>1;
if(cx<=mid) recovey(x<<1, l, mid, cx, pos, d+1);
else recovey(x<<1|1, mid+1, r, cx, pos, d+1);
}
void dfs(int x, int d)
{
if(x!=1)
{
int limit=dep[d]-l[x], nxt=v[query(1, 1, n, find(d-1, limit), p[x])];
f[x]=f[nxt]+(deep[x]-deep[nxt])*p[x]+q[x];
}
update(1, 1, n, d, x, 1);
for(int i=last[x], too;i;i=e[i].pre)
deep[too=e[i].too]=deep[x]+e[i].dis, dep[d+1]=dep[d]+e[i].dis, dfs(too, d+1);
recovey(1, 1, n, d, x, 1);
}
#undef int
int main()
{
read(n); read(m);
for(int i=2;i<=n;i++)
read(x), read(z), read(p[i]), read(q[i]), read(l[i]), add(x, i, z);
sum=1; build(1, 1, n); dfs(1, 1);
for(int i=2;i<=n;i++) printf("%lld\n", f[i]);
}