BZOJ_1803_Spoj1487 Query on a tree III_主席树+dfs序

BZOJ_1803_Spoj1487 Query on a tree III_主席树

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

恶心的是需要输出编号，而再开一个来存主席树上节点对应树上节点编号就非常卡空间。

于是把权值离散化。

代码：

#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)]);
    }
}

原文地址：https://www.cnblogs.com/suika/p/8967853.html