【Codeforces Round 326 (Div 2)E】【树链剖分】Duff in the Army 树上给定路径上编号最小的几个人
Recently Duff has been a soldier in the army. Malek is her commander.
Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.
There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is iand he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.
Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.
To answer a query:
Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order.
If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).
Duff is very busy at the moment, so she asked you to help her and answer the queries.
The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).
The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).
Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).
Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).
For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.
5 4 5 1 3 1 2 1 4 4 5 2 1 4 3 4 5 6 1 5 2 5 5 10 2 3 3 5 3 1
1 3 2 2 3 0 3 1 2 4 1 2
Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):
#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=1e5+10,M=0,Z=1e9+7,ms63=1061109567;
int casenum,casei;
int n,m,q,tim;
int x,y,w;
struct A
{
int l,r;
int g;
int b[10];
}c[1<<18];
vector<int>a[N];
int g[N];
int b[N][10];
int G,tmpg;
int B[10],tmpb[10];
int fa[N],son[N],dep[N],size[N],pos[N],rev[N],top[N];
void dfs1(int x)
{
size[x]=1;
son[x]=0;
for(int i=a[x].size()-1;~i;--i)
{
int y=a[x][i];
if(y==fa[x])continue;
fa[y]=x;
dep[y]=dep[x]+1;
dfs1(y);
size[x]+=size[y];
if(size[y]>size[son[x]])son[x]=y;
}
}
void dfs2(int x,int chain)
{
pos[x]=++tim;
rev[tim]=x;
top[x]=chain;
if(son[x]==0)return;
dfs2(son[x],chain);
for(int i=a[x].size()-1;~i;--i)
{
int y=a[x][i];
if(y!=fa[x]&&y!=son[x])dfs2(y,y);
}
}
void merge(int &g,int &gl,int &gr,int b[],int bl[],int br[])
{
int p=0,pl=0,pr=0;
while(p<10)
{
if(pl<gl&&pr<gr)
{
if(bl[pl]<br[pr])b[p++]=bl[pl++];
else b[p++]=br[pr++];
}
else if(pl<gl)b[p++]=bl[pl++];
else if(pr<gr)b[p++]=br[pr++];
else break;
}
g=p;
}
void pushup(int o)
{
merge(c[o].g,c[ls].g,c[rs].g,c[o].b,c[ls].b,c[rs].b);
}
void build(int o,int l,int r)
{
c[o].l=l;
c[o].r=r;
if(l==r)
{
int v=rev[l];
c[o].g=g[v];
MC(c[o].b,b[v]);
return;
}
int m=(l+r)>>1;
build(ls,l,m);
build(rs,m+1,r);
pushup(o);
}
void update(int o,int l,int r)
{
if(c[o].l==l&&c[o].r==r)
{
tmpg=G;
MC(tmpb,B);
merge(G,tmpg,c[o].g,B,tmpb,c[o].b);
return;
}
int m=(c[o].l+c[o].r)>>1;
if(r<=m)update(ls,l,r);
else if(l>m)update(rs,l,r);
else
{
update(ls,l,m);
update(rs,m+1,r);
}
}
void SOLVE(int x,int y,int w)
{
G=0;
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
update(1,pos[top[x]],pos[x]);
x=fa[top[x]];
}
if(pos[x]>pos[y])swap(x,y);
update(1,pos[x],pos[y]);
gmin(G,w);
printf("%d",G);
for(int i=0;i<G;i++)printf(" %d",B[i]);
puts("");
}
int main()
{
while(~scanf("%d%d%d",&n,&m,&q))
{
for(int i=1;i<=n;i++)
{
a[i].clear();
g[i]=0;
}
for(int i=1;i<n;i++)
{
scanf("%d%d",&x,&y);
a[x].push_back(y);
a[y].push_back(x);
}
for(int i=1;i<=m;i++)
{
scanf("%d",&x);
if(g[x]<10)b[x][g[x]++]=i;
}
dep[1]=0;fa[1]=0;dfs1(1);
tim=0;dfs2(1,1);
build(1,1,n);
for(int i=1;i<=q;i++)
{
scanf("%d%d%d",&x,&y,&w);
SOLVE(x,y,w);
}
}
return 0;
}
/*
【trick&&吐槽】
这题明明要比D简单，是裸的树链剖分。
我写得很慢慢悠悠，然而竟然是div2前几个做出来的，太吃惊了！
【题意】
给你一棵有n（1e5）个节点树，
然后有m（1e5）个人，告诉你编号为i个人位于什么节点。
有q（1e5）个询问，问你对于（x,y）路径上的点中，最小编号的w（[1,10]）个人的编号。（当然如果没有w个人，有几个输出几个）
【类型】
树链剖分
【分析】
因为w实在是太小了，所以每个节点直接记录以它为父节点的最小编号的10个人的编号即可。
用线段树，通过类似于归并排序的方式维护一下。然后因为是在树上，套一个树链剖分的模板就可以AC了。
因为没有修改，所以线段树只用pushup，也就写个合并操作即可。
【时间复杂度&&优化】
O（（n+q）logn*10）
*/