Query on a tree 树链剖分

题意:

给你一棵树，和树上边的权值，在有q组询问a,b，问你从节点a->节点1的路径上，不小于b的最大的边的权值是多少，输出

离线维护最大值线段树即可

模板题

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define see(x) (cerr<<(#x)<<‘=‘<<(x)<<endl)
#define pb push_back
#define lson l,m,pos<<1
#define rson m+1,r,pos<<1|1
#define inf 0x3f3f3f3f
#define CLR(A,v)  memset(A,v,sizeof A)
typedef pair<int,int>pii;
//////////////////////////////////
const int N=1e5+10;

int t[N<<2],n,m;
void up(int pos)
{
    t[pos]=max(t[pos<<1],t[pos<<1|1]);
}
void build(int l,int r,int pos)
{
    if(l==r){t[pos]=0;return ;}
    int m=(l+r)>>1;
    build(lson);build(rson);up(pos);
}
void upnode(int x,int v,int l,int r,int pos)
{
    if(l==r){t[pos]=v;return ;}
    int m=(l+r)>>1;
    if(x<=m)upnode(x,v,lson);
    else upnode(x,v,rson);
    up(pos);
}
int qmax(int L,int R,int l,int r,int pos)
{
    int ans=-inf;
    if(L<=l&&r<=R)return t[pos];int m=(l+r)>>1;
    if(L<=m)ans=max(ans,qmax(L,R,lson));
    if(R>m)ans=max(ans,qmax(L,R,rson));
    up(pos);return ans;
}
int id[N],head[N],pos,cnt,top[N],fa[N],son[N],siz[N],dep[N],a,b,c;
struct Edge
{
    int to,nex;
}edge[N<<1];
void add(int a,int b)
{
    edge[++pos]=Edge{b,head[a]};
    head[a]=pos;
}
void dfs1(int x,int f)
{
    fa[x]=f;dep[x]=dep[f]+1;siz[x]=1;son[x]=0;
    for(int i=head[x];i;i=edge[i].nex)
    {
        int v=edge[i].to;
        if(v==f)continue;
        dfs1(v,x);
        siz[x]+=siz[v];
        if(siz[son[x]]<siz[v])son[x]=v;
    }
}
void dfs2(int x,int topf)
{
    id[x]=++cnt;top[x]=topf;
    if(son[x])dfs2(son[x],topf);
    for(int i=head[x];i;i=edge[i].nex)
    {
        int v=edge[i].to;
        if(v==son[x]||v==fa[x])continue;
        dfs2(v,v);
    }
}
int Qmax(int x,int y)
{
    int ans=-inf;
    while(top[x]!=top[y])
    {
        if(dep[top[x]]<dep[top[y]])swap(x,y);
        ans=max(ans,qmax(id[top[x]],id[x],1,n,1));
        x=fa[top[x]];
    }
    if(dep[x]>dep[y])swap(x,y);

    ans=max(ans,qmax(id[x]+1,id[y],1,n,1));
    return ans;
}
void init()
{
    pos=cnt=0;CLR(head,0);dep[1]=siz[0]=0;
}
struct node
{
    int x,y,id;
}s[N];
bool cmp(node a,node b)
{
    return a.y<b.y;
}
struct Edge2
{
    int u,v,w;
}edge2[N];
bool cmp2(Edge2 a,Edge2 b)
{
    return a.w<b.w;
}
int ans[N];
int main()
{
    int cas;RI(cas);
    while(cas--)
    {
        RI(n);init();
        rep(i,1,n-1)
        {
            RIII(edge2[i].u,edge2[i].v,edge2[i].w);add(edge2[i].u,edge2[i].v);add(edge2[i].v,edge2[i].u);
        }
        dfs1(1,1);dfs2(1,1);build(1,n,1);

        RI(m);
        rep(i,1,m)RII(s[i].x,s[i].y),s[i].id=i;
        sort(s+1,s+1+m,cmp);sort(edge2+1,edge2+1+n-1,cmp2);

        int L=1;
        rep(i,1,m)
        {
            while(edge2[L].w<=s[i].y&&L<=n-1 )
            {
                int u=edge2[L].u,v=edge2[L].v;
                if(dep[u]<dep[v])upnode(id[v],edge2[L].w,1,n,1);
                else upnode(id[u],edge2[L].w,1,n,1);L++;
            }
            ans[s[i].id]=Qmax(1,s[i].x);
        }
        rep(i,1,m)
        printf("%d\n",ans[i]==0||ans[i]==-inf?-1:ans[i]);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/bxd123/p/11175345.html