主席树+LCA+spoj10628

10628. Count on a tree

Problem code: COT

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 5105 2 9 3 8 5 7 71 2        1 31 43 53 63 74 82 5 12 5 22 5 32 5 47 8 2 

Output:2891057 

题意：给出一棵树，询问两点之间第k小的权值是多少

思路：主席树记录从根节点到当前节点出现在1..m（权值按从小到大排序后）中的数（跟poj2104求区间第k小的数一样，这里只不过是书上从根到当前节点），这样当查询时候

在线求出u,v的lca，然后就跟查询区间第k小的值一样了

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=200010;const int maxm=maxn*50;int N,M,m,num;int w[maxn];int order[maxn],cnt[maxm],lson[maxm],rson[maxm],T[maxm];int E[maxn*2],dep[maxn*2],pos[maxn],vis[maxn];int d[maxn*2][20];int head[maxn];struct node{    int v,next;}edge[maxn*2];void add_edge(int u,int v){    edge[num].v=v;    edge[num].next=head[u];    head[u]=num++;}void init(){    num=0;    for(int i=0;i<=N;i++)    {        vis[i]=0;        pos[i]=head[i]=-1;    }    for(int i=1;i<=N;i++)order[i]=w[i];    sort(order+1,order+1+N);    m=unique(order+1,order+1+N)-order-1;}//LCA部分void LCA_dfs(int u,int depth){    E[++num]=u,dep[num]=depth;    if(pos[u]==-1)pos[u]=num;    vis[u]=1;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(vis[v])continue;        LCA_dfs(v,depth+1);        E[++num]=u;        dep[num]=depth;    }}void initRMQ(int n){    for(int i=0;i<=n;i++)d[i][0]=i;    for(int j=1;(1<<j)<=n;j++)    {        for(int i=1;i+(1<<j)<=n;i++)        {            int x=d[i][j-1],y=d[i+(1<<(j-1))][j-1];            if(dep[x]<=dep[y])d[i][j]=x;            else d[i][j]=y;        }    }}int LCA_query(int u,int v){    int x=pos[u],y=pos[v];    if(x>y)swap(x,y);    int k=0;    while((1<<(k+1))<=y-x+1)k++;    int l=d[x][k],r=d[y-(1<<k)+1][k];    if(dep[l]<=dep[r])return E[l];    else return E[r];}//主席树部分int  build(int l,int r){    int root=num++;    cnt[root]=0;    if(l!=r)    {        int mid=(l+r)>>1;        lson[root]=build(l,mid);        rson[root]=build(mid+1,r);    }    return root;}int update(int root,int pos,int val){    int newroot=num++,tmp=newroot;    int l=1,r=m;    cnt[newroot]=cnt[root]+val;    while(l<r)    {        int mid=(l+r)>>1;        if(pos<=mid)        {            r=mid;            lson[newroot]=num++,rson[newroot]=rson[root];            newroot=lson[newroot],root=lson[root];        }        else        {            l=mid+1;            rson[newroot]=num++,lson[newroot]=lson[root];            newroot=rson[newroot],root=rson[root];        }        cnt[newroot]=cnt[root]+val;    }    return tmp;}int find(int x){    return lower_bound(order+1,order+1+m,x)-order;}void dfs_build(int u,int fa){    int pos=find(w[u]);    T[u]=update(T[fa],pos,1);    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==fa)continue;        dfs_build(v,u);    }}int query(int left,int right,int lca,int k){    int l=1,r=m;    int pos=find(w[lca]);    lca=T[lca];    while(l<r)    {        int mid=(l+r)>>1;        //注意减掉两倍的lca后，相当于lca这个点没有了，要判断是不是应该加上        int tmp=cnt[lson[left]]+cnt[lson[right]]-2*cnt[lson[lca]]+(pos>=l&&pos<=mid);        if(tmp>=k)        {            r=mid;            left=lson[left];            right=lson[right];            lca=lson[lca];        }        else        {            l=mid+1;            k-=tmp;            left=rson[left];            right=rson[right];            lca=rson[lca];        }    }    return l;}int main(){    while(scanf("%d%d",&N,&M)!=EOF)    {        for(int i=1;i<=N;i++)            scanf("%d",&w[i]);        init();        for(int i=1;i<N;i++)        {            int u,v;            scanf("%d%d",&u,&v);            add_edge(u,v);            add_edge(v,u);        }        num=0;        LCA_dfs(1,1);        initRMQ(num);        num=0;        T[0]=build(1,m);        dfs_build(1,0);        while(M--)        {            int u,v,k;            scanf("%d%d%d",&u,&v,&k);            int lca=LCA_query(u,v);            printf("%d\n",order[query(T[u],T[v],lca,k)]);        }    }    return 0;}