SPOJ-COT-Count on a tree

COT-Count on a tree

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

题目大意：树上任意两点之间路径中的第k小
解题思路：主席树+LCA

#include<iostream>#include<cstdio>#include<vector>#include<algorithm>#include<map>#include<cmath>#include<set>#include<cstring>using namespace std;typedef long long LL;const int MAXN=1e5+5;int n,m,tot;int cnt,head[MAXN];int len,root[MAXN],lson[MAXN*20],rson[MAXN*20],val[MAXN*20];int num,dep[MAXN*2],ver[MAXN*2],fst[MAXN],dp[MAXN*2][20],fa[MAXN];bool vis[MAXN];int a[MAXN],b[MAXN];struct Edge{    int to,nxt;}e[MAXN*2];void addedge(int u,int v){    e[cnt].to=v;    e[cnt].nxt=head[u];    head[u]=cnt++;}void build(int l,int r,int &rt){    rt=++tot;    val[rt]=0;    if(l==r) return;    int mid=(l+r)>>1;    build(l,mid,lson[rt]);    build(mid+1,r,rson[rt]);}void update(int pre,int &rt,int l,int r,int v){    rt=++tot;    lson[rt]=lson[pre];rson[rt]=rson[pre];val[rt]=val[pre]+1;    if(l==r) return;    int mid=(l+r)>>1;    if(v<=mid) update(lson[pre],lson[rt],l,mid,v);    else update(rson[pre],rson[rt],mid+1,r,v);}void dfs(int u,int fat,int d){    vis[u]=true;ver[++num]=u;dep[num]=d;fst[u]=num;fa[u]=fat;    update(root[fat],root[u],1,len,a[u]);    for(int i=head[u];i!=-1;i=e[i].nxt)    {        int to=e[i].to;        if(!vis[to])        {            dfs(to,u,d+1);            ver[++num]=u;dep[num]=d;        }    }}void ST(int n){    for(int i=1;i<=n;i++) dp[i][0]=i;    for(int j=1;(1<<j)<=n;j++)    {        for(int i=1;i<=n-(1<<j)+1;i++)        {            if(dep[dp[i][j-1]]<dep[dp[i+(1<<(j-1))][j-1]])            dp[i][j]=dp[i][j-1];            else dp[i][j]=dp[i+(1<<(j-1))][j-1];        }    }}int RMQ(int l,int r){    int k=log(r-l+1)/log(2);    if(dep[dp[l][k]]<dep[dp[r-(1<<k)+1][k]]) return dp[l][k];    else return dp[r-(1<<k)+1][k];}int LCA(int u,int v){    u=fst[u],v=fst[v];    if(u>v) swap(u,v);    int res=RMQ(u,v);    return ver[res];}int query(int ss,int tt,int lca,int lcafa,int l,int r,int k){    if(l==r) return l;    int mid=(l+r)>>1;    int tmp=val[lson[ss]]+val[lson[tt]]-val[lson[lca]]-val[lson[lcafa]];    if(k<=tmp) return query(lson[ss],lson[tt],lson[lca],lson[lcafa],l,mid,k);    else return query(rson[ss],rson[tt],rson[lca],rson[lcafa],mid+1,r,k-tmp);}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b[i]=a[i];        }        sort(b+1,b+1+n);        len=unique(b+1,b+1+n)-(b+1);        for(int i=1;i<=n;i++) a[i]=lower_bound(b+1,b+len+1,a[i])-b;        cnt=0;        memset(head,-1,sizeof(head));        int u,v,k;        for(int i=1;i<=n-1;i++)        {            scanf("%d%d",&u,&v);            addedge(u,v);addedge(v,u);        }        tot=0;        build(1,len,root[0]);        num=0;        memset(vis,false,sizeof(vis));        dfs(1,0,1);        ST(2*n-1);        int lca;        for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&k);            lca=LCA(u,v);            printf("%d\n",b[query(root[u],root[v],root[lca],root[fa[lca]],1,len,k)]);        }    }    return 0;}