描述

http://www.spoj.com/problems/COT/

给出一棵n个节点的树,树上每一个节点有权值.m次询问,求书上u,v路径中第k小的权值.

COT - Count on a tree

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You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight.

We will ask you to perform the following operation:

u v k : ask for the kth minimum weight on the path from node u to node v

Input

In the first line there are two integers N and M.(N,M<=100000)

In the second line there are N integers.The ith integer denotes the weight of the ith node.

In the next N-1 lines,each line contains two integers u v,which describes an edge (u,v).

In the next M lines,each line contains three integers u v k,which means an operation asking for the kth minimum weight on the path from node u to node v.

Output

For each operation,print its result.

Example

Input:

8 5

8 5

105 2 9 3 8 5 7 7

1 2

1 3

1 4

3 5

3 6

3 7

4 8
2 5 1
2 5 2
2 5 3
2 5 4
7 8 2

Output:

2
8
9
105
7

分析

POJ_2104_Kth(主席树)

现在是把原来的问题搬到树上去了.首先我们肯定要求lca,新学了Tarjan的离线算法.

每一个点建立到根节点的主席树,这样最后的结果就是u+v-2*lca,如果lca在所要求的区间内,还要再加上lca.(或者u+v-lca-p[lca]).

自己理解一下吧...挺简单的.

 #include <cstdio>

 #include <algorithm>

 #include <vector>

 using namespace std;

 const int maxn=+;

 int n,m,cnt,num;

 int a[maxn],id[maxn],b[maxn],head[maxn],p[maxn],f[maxn],root[maxn];

 bool vis[maxn];

 struct edge{

     int to,next;

     edge(){}

     edge(int a,int b):to(a),next(b){}

 }g[maxn<<];

 struct Qry{

     int u,v,k,lca;

     Qry(){}

     Qry(int a,int b,int c,int d):u(a),v(b),k(c),lca(d){}

 }Q[maxn];

 struct node{ int l,r,s; }t[maxn*];

 struct qry{

     int v,id;

     qry(){}

     qry(int a,int b):v(a),id(b){}

 };

 vector <qry> q[maxn];

 inline int find(int x){ return x==f[x]?x:f[x]=find(f[x]); }

 void add_edge(int u,int v){

     g[++cnt]=edge(v,head[u]); head[u]=cnt;

     g[++cnt]=edge(u,head[v]); head[v]=cnt;

 }

 void update(int l,int r,int &pos,int d){

     t[++num]=t[pos]; pos=num; t[pos].s++;

     if(l==r) return;

     int mid=l+(r-l)/;

     if(d<=mid) update(l,mid,t[pos].l,d);

     else update(mid+,r,t[pos].r,d);

 }

 bool cmp(int x,int y){ return a[x]<a[y]; }

 void dfs(int u){

     f[u]=u; root[u]=root[p[u]]; update(,n,root[u],b[u]);

     for(int i=head[u];i;i=g[i].next){

         if(g[i].to!=p[u]){

             p[g[i].to]=u;

             dfs(g[i].to);

             f[g[i].to]=u;

         }

     }

     vis[u]=true;

     int size=q[u].size();

     for(int i=;i<size;i++) if(vis[q[u][i].v]) Q[q[u][i].id].lca=find(q[u][i].v);

 }

 void init(){

     scanf("%d%d",&n,&m);

     for(int i=;i<=n;i++) scanf("%d",&a[i]), id[i]=i;

     sort(id+,id+n+,cmp);

     for(int i=;i<=n;i++) b[id[i]]=i;

     for(int i=;i<n;i++){

         int u,v;

         scanf("%d%d",&u,&v);

         add_edge(u,v);

     }

     for(int i=;i<=m;i++){

         scanf("%d%d%d",&Q[i].u,&Q[i].v,&Q[i].k);

         q[Q[i].u].push_back(qry(Q[i].v,i)); q[Q[i].v].push_back(qry(Q[i].u,i));

     }

 }

 int query(int l,int r,int x,int y,int ra,int a,int k){

     if(l==r) return l;

     int mid=l+(r-l)/;

     int s=t[t[x].l].s+t[t[y].l].s-*t[t[ra].l].s;

     if(b[a]>=l&&b[a]<=mid) s++;

     if(k<=s) return query(l,mid,t[x].l,t[y].l,t[ra].l,a,k);

     else return query(mid+,r,t[x].r,t[y].r,t[ra].r,a,k-s);

 }

 void solve(){

     dfs();

     for(int i=;i<=m;i++){

         if(Q[i].u==Q[i].v){ printf("%d\n",a[Q[i].u]); continue; }

         printf("%d\n",a[id[query(,n,root[Q[i].u],root[Q[i].v],root[Q[i].lca],Q[i].lca,Q[i].k)]]);

     }

 }

 int main(){

     init();

     solve();

     return ;

 }

巴特西

SPOJ_10628_Count_on_a_Tree_(主席树+Tarjan)

描述

COT - Count on a tree

Input

Output

Example

分析

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