BZOJ_1803_Spoj1487 Query on a tree III_Chairman tree+dfs sequence

BZOJ_1803_Spoj1487 Query on a tree III_Chairman tree

Description

You are given a node-labeled rooted tree with n nodes. Define the query (x, k): Find the node whose label is k-th largest in the subtree of the node x. Assume no two nodes have the same labels.

Input

The first line contains one integer n (1 <= n <= 10^5). The next line contains n integers li (0 <= li <= 109) which denotes the label of the i-th node. Each line of the following n - 1 lines contains two integers u, v. They denote there is an edge between node u and node v. Node 1 is the root of the tree. The next line contains one integer m (1 <= m <= 10^4) which denotes the number of the queries. Each line of the next m contains two integers x, k. (k <= the total node number in the subtree of x)

Output

For each query (x, k), output the index of the node whose label is the k-th largest in the subtree of the node x.

Sample Input

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

Sample Output

5
4
5

---

 The kth subtree is the smallest, and the chairman tree is solved on the tree.

The disgusting thing is that it needs to output the number, and opening another one to store the node number on the chairman tree corresponding to the node number on the tree is very space-consuming.

So the weights are discretized.

 

Code:

#include <cstdio>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 100050
#define maxn n
int head[N],to[N<<1],nxt[N<<1],cnt,n,m,S[N],dfn[N],root[N],t[N*30],ls[N*30],rs[N*30],tot,val[N],son[N],rr[N];
inline void add(int u,int v) {
    to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt;
}
struct A {
    int num, id,v;
}a[N];
bool cmp1(const A &x,const A &y){return x.num<y.num;}
bool cmp2(const A &x,const A &y){return x.id<y.id;}
void dfs(int x,int y) {
    int i;
    S[++S[0]]=x; dfn[x]=S[0];
    for(i=head[x];i;i=nxt[i]) {
        if(to[i]!=y) {
            dfs(to[i],x);
        }
    }
    son[x]=S[0];
}
void insert(int &y,int x,int l,int r,int v) {
    y=++tot; t[y]=t[x]+1;
    if(l==r) return ;
    int mid=(l+r)>>1;
    if(v<=mid) rs[y]=rs[x],insert(ls[y],ls[x],l,mid,v);
    else ls[y]=ls[x],insert(rs[y],rs[x],mid+1,r,v);
}
int query (int x, int y, int l, int r, int k) {
    if(l==r) return l;
    int mid=(l+r)>>1,sizls=t[ls[x]]-t[ls[y]];
    if(k<=sizls) return query(ls[x],ls[y],l,mid,k);
    else return query(rs[x],rs[y],mid+1,r,k-sizls);
}
int main() {
    scanf("%d",&n);
    int i,x,y,k;
    for(i=1;i<=n;i++) scanf("%d",&a[i].num),a[i].id=i;
    sort(a+1,a+n+1,cmp1);
    int j=0;a[0].num=43345;
    for(i=1;i<=n;i++) {
        if(a[i].num!=a[i-1].num) j++;
        a[i].v=j; rr[j]=a[i].id;
    }
    sort(a+1,a+n+1,cmp2);
    for(i=1;i<n;i++) {
        scanf("%d%d",&x,&y); add(x,y); add(y,x);
    }
    dfs(1,0);
    for(i=1;i<=n;i++) {
        insert(root[i],root[i-1],0,maxn,a[S[i]].v);
    }
    scanf("%d",&m);
    while(m--) {
        scanf("%d%d",&x,&k);
        printf("%d\n",rr[query(root[son[x]],root[dfn[x]-1],0,maxn,k)]);
    }
}