BZOJ1803: Spoj1487 Query on a tree III

1803: Spoj1487 Query on a tree III

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 286  Solved: 125
[Submit][Status]

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

题解：

说好的第k大呢，人与人之间的信任呢。。。

子树查询  dfs序+主席树搞掉。。。

代码：

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#define inf 1000000000

#define maxn 200000+5

#define maxm 3000000+5

#define eps 1e-10

#define ll long long

#define pa pair<int,int>

#define for0(i,n) for(int i=0;i<=(n);i++)

#define for1(i,n) for(int i=1;i<=(n);i++)

#define for2(i,x,y) for(int i=(x);i<=(y);i++)

#define for3(i,x,y) for(int i=(x);i>=(y);i--)

#define mod 1000000007

using namespace std;

inline int read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}

    return x*f;

}
struct edge{int go,next;}e[2*maxn];
int n,m,cnt,tot,a[maxn],b[maxn],c[maxn],d[maxn],rt[maxn],ls[maxm],rs[maxm],s[maxm];
int head[maxn],l[maxn],r[maxn],t[maxn][2];
inline void insert(int x,int y)
{
    e[++tot]=(edge){y,head[x]};head[x]=tot;
    e[++tot]=(edge){x,head[y]};head[y]=tot;
}
inline bool cmp(int x,int y){return a[x]<a[y];}
inline void update(int l,int r,int x,int &y,int z)
{
    y=++cnt;
    s[y]=s[x]+1;
    if(l==r)return;
    ls[y]=ls[x];rs[y]=rs[x];
    int mid=(l+r)>>1;
    if(z<=mid)update(l,mid,ls[x],ls[y],z);else update(mid+1,r,rs[x],rs[y],z);
}
inline void dfs(int x,int f)
{
    t[x][0]=++m;
    update(1,n,rt[m-1],rt[m],c[x]);
    for(int i=head[x];i;i=e[i].next)if(e[i].go!=f)dfs(e[i].go,x);
    t[x][1]=m;
}

int main()

{

    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);

    n=read();
    for1(i,n)a[i]=read(),b[i]=i;
    sort(b+1,b+n+1,cmp);
    for1(i,n)c[b[i]]=i;
    for1(i,n-1)insert(read(),read());
    dfs(1,0);
    m=read();
    while(m--)
    {
        int x=read(),k=read(),l=1,r=n,xx=rt[t[x][0]-1],yy=rt[t[x][1]];
        //k=s[yy]-s[xx]+1-k;
        while(l!=r)
        {
           int mid=(l+r)>>1,t=s[ls[yy]]-s[ls[xx]];
           if(t>=k){xx=ls[xx];yy=ls[yy];r=mid;}
           else {xx=rs[xx];yy=rs[yy];l=mid+1;k-=t;}
        }
        printf("%d\n",b[l]);
    }

    return 0;

}