BZOJ 1803: Spoj1487 Query on a tree III 主席树题解

Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 595 Solved: 268

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

5

---

显然主席树是可以瞎搞的，只需要跑个dfs序，然后在dfs序上建主席树，然后就可以了嘛每次查一个点in和out中间的信息即可

---

#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>#include<map>#include<queue>#include<set>const int MAXN=111111;using namespace std;struct Line{    int to,nxt;}line[MAXN*2+100];struct Tree{    int lson,rson,sum;}tree[MAXN*40];int head[MAXN],tail,sz;map<int,int>mm;void add_line(int from,int to){    line[++tail].to=to;    line[tail].nxt=head[from];    head[from]=tail;}int timer,a[MAXN*2],root[MAXN*2],in[MAXN*2],out[MAXN*2],n,m,x,kk,from,to,dfn[MAXN*2];vector<int>v;int id(int x){return int(lower_bound(v.begin(),v.end(),x)-v.begin()+1);}void dfs(int u,int fa){    in[u]=++timer;    dfn[in[u]]=u;    for(int i=head[u];i;i=line[i].nxt){        int v=line[i].to;        if(v!=fa) dfs(v,u);    }    out[u]=timer;}void modify(int x,int &y,int l,int r,int loc){    tree[++sz]=tree[x];y=sz;tree[y].sum++;    if(l==r) return;    int mid=(l+r)>>1;    if(loc<=mid) modify(tree[x].lson,tree[y].lson,l,mid,loc);    else         modify(tree[x].rson,tree[y].rson,mid+1,r,loc);}int query(int x,int y,int k,int l,int r){    if(l==r)return l;    int mid=(l+r)>>1;    int t=tree[tree[y].lson].sum-tree[tree[x].lson].sum;    if(k<=t) return query(tree[x].lson,tree[y].lson,k,l,mid);    else     return query(tree[x].rson,tree[y].rson,k-t,mid+1,r);}int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",&a[i]),v.push_back(a[i]),mm[a[i]]=i;    for(int i=1;i<=n-1;i++){scanf("%d%d",&from,&to);add_line(from,to);add_line(to,from);}    dfs(1,0);    sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end());    for(int i=1;i<=n;i++) modify(root[i-1],root[i],1,n,id(a[dfn[i]]));    scanf("%d",&m);    while(m--){        scanf("%d%d",&x,&kk);        int vall=v[query(root[in[x]-1],root[out[x]],kk,1,n)-1];        printf("%d\n",mm[vall]);    }    return 0;}