题意
显然树剖套李超树。
考虑怎么算函数值：
设((x,y))的(lca)为(z)，我们插一条斜率为(k)，截距为(b)的线段。
((x,z))上的点(u)：
(f(u)=k*(dis[x]-dis[u])+b=-k*dis[u]+(k*dis[x]+b))
所以对这条路径插入斜率为(-k)，截距为(k*dis[x]+b)的线段
((y,z))上的点(u)：
(f(u)=k*(dis[y]+dis[u]-2*dis[z])+b=k*dis[u]+k*(dis[y]-2*dis[z])+b)
同理
code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ls(p) (p<<1)
#define rs(p) (p<<1|1)
#define minn(p)(seg[p].minn)
#define minid(p) (seg[p].minid)
const int maxn=100010;
const int inf=123456789123456789ll;
int n,m,cnt,tim;
int head[maxn],dep[maxn],dfn[maxn],pos[maxn],size[maxn],son[maxn],pre[maxn],top[maxn],dis[maxn];
struct edge{int to,nxt,dis;}e[maxn<<1];
struct Line
{
int k,b;
inline int calc(int x){return k*x+b;}
};
struct Seg{int minn;Line minid;}seg[maxn<<2];
inline int read()
{
char c=getchar();int res=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9')res=res*10+c-'0',c=getchar();
return res*f;
}
inline void add(int u,int v,int w)
{
e[++cnt].nxt=head[u];
head[u]=cnt;
e[cnt].to=v;
e[cnt].dis=w;
}
void dfs1(int x,int fa)
{
pre[x]=fa;dep[x]=dep[fa]+1;size[x]=1;
for(int i=head[x];i;i=e[i].nxt)
{
int y=e[i].to;
if(y==fa)continue;
dis[y]=dis[x]+e[i].dis;
dfs1(y,x);size[x]+=size[y];
if(size[son[x]]<size[y])son[x]=y;
}
}
void dfs2(int x,int tp)
{
top[x]=tp;dfn[x]=++tim;pos[tim]=x;
if(son[x])dfs2(son[x],tp);
for(int i=head[x];i;i=e[i].nxt)
{
int y=e[i].to;
if(y==pre[x]||y==son[x])continue;
dfs2(y,y);
}
}
inline int lca(int x,int y)
{
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
x=pre[top[x]];
}
if(dep[x]>dep[y])swap(x,y);
return x;
}
inline void up(int p){minn(p)=min(minn(p),min(minn(ls(p)),minn(rs(p))));}
void build(int p,int l,int r)
{
minn(p)=inf;minid(p)=(Line){0,inf};
if(l==r)return;
int mid=(l+r)>>1;
build(ls(p),l,mid);build(rs(p),mid+1,r);
}
inline void move(int p,int l,int r,Line id)
{
int nowl=minid(p).calc(dis[pos[l]]),nowr=minid(p).calc(dis[pos[r]]);
int tmpl=id.calc(dis[pos[l]]),tmpr=id.calc(dis[pos[r]]);
if(nowl<=tmpl&&nowr<=tmpr)return;
if(nowl>=tmpl&&nowr>=tmpr)
{
minid(p)=id;minn(p)=min(minn(p),min(tmpl,tmpr));
return;
}
double point=1.0*(minid(p).b-id.b)/(1.0*(id.k-minid(p).k));
int mid=(l+r)>>1;
if(tmpl>nowl)
{
if(point<=(double)dis[pos[mid]])move(ls(p),l,mid,minid(p)),minid(p)=id;
else move(rs(p),mid+1,r,id);
}
else
{
if(point<=(double)dis[pos[mid]])move(ls(p),l,mid,id);
else move(rs(p),mid+1,r,minid(p)),minid(p)=id;
}
minn(p)=min(minn(p),min(tmpl,tmpr));
up(p);
}
void change(int p,int l,int r,int ql,int qr,Line id)
{
if(l>=ql&&r<=qr){move(p,l,r,id);return;}
int mid=(l+r)>>1;
if(ql<=mid)change(ls(p),l,mid,ql,qr,id);
if(qr>mid)change(rs(p),mid+1,r,ql,qr,id);
up(p);
}
int query(int p,int l,int r,int ql,int qr)
{
if(l>=ql&&r<=qr)return minn(p);
int mid=(l+r)>>1,res=min(minid(p).calc(dis[pos[max(l,ql)]]),minid(p).calc(dis[pos[min(r,qr)]]));
if(ql<=mid)res=min(res,query(ls(p),l,mid,ql,qr));
if(qr>mid)res=min(res,query(rs(p),mid+1,r,ql,qr));
return res;
}
inline void trchange(int x,int y,int k,int b)
{
while(top[x]!=top[y])
{
change(1,1,n,dfn[top[x]],dfn[x],(Line){k,b});
x=pre[top[x]];
}
change(1,1,n,dfn[y],dfn[x],(Line){k,b});
}
inline void trsolve(int x,int y,int k,int b)
{
int z=lca(x,y);
trchange(x,z,-k,k*dis[x]+b);
trchange(y,z,k,k*(dis[x]-(dis[z]<<1))+b);
}
inline int trquery(int x,int y)
{
int res=inf;
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
res=min(res,query(1,1,n,dfn[top[x]],dfn[x]));
x=pre[top[x]];
}
if(dep[x]>dep[y])swap(x,y);
res=min(res,query(1,1,n,dfn[x],dfn[y]));
return res;
}
signed main()
{
//freopen("test.in","r",stdin);
//freopen("test.out","w",stdout);
n=read(),m=read();
for(int i=1;i<n;i++)
{
int u=read(),v=read(),w=read();
add(u,v,w),add(v,u,w);
}
dfs1(1,0),dfs2(1,1);
build(1,1,n);
for(int i=1;i<=m;i++)
{
int op=read(),x=read(),y=read(),k,b;
if(op==1)k=read(),b=read(),trsolve(x,y,k,b);
else printf("%lld
",trquery(x,y));
}
return 0;
}