bzoj 1803(主席树+dfs序)

1803: Spoj1487 Query on a tree III

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 510  Solved: 221

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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

解题思路：主席树模板+dfs序。



#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

using namespace std;

struct ss

 {

  int zhi,q;

 }a[200011];

int len,tail=0,cnt=0,opp=0,n;

int to[210000],next[210000],h[210000];

int qg[210000],st[200011],en[200011];

int cow[200100],dui[200100],b[200100];

int sum[4010000],lx[4010000],rx[4010000];

inline int read()

{
char y; int x=0,f=1; y=getchar();
while (y<'0' || y>'9') {if (y=='-')f=-1; y=getchar();}
while (y>='0' && y<='9') {x=x*10+int(y)-48; y=getchar();}
return x*f;

}

bool cmp(ss x,ss y)

 {

  return x.zhi<y.zhi;

 }

void insert(int x,int y)

 {

  ++len; to[len]=y; next[len]=h[x]; h[x]=len;

 }

void dfs(int now,int f)

 {

   ++tail; qg[tail]=now; st[now]=tail;

   int u=h[now];

   while (u!=0)

    {

    if (to[u]!=f)

    {

    dfs(to[u],now);
}

    u=next[u];
}

   ++tail; qg[tail]=now; en[now]=tail;

 }

void updata(int l,int r,int x,int &y,int og)

 {

  y=++opp; sum[y]=sum[x]+1;

  if (l==r)return;

  int mid=(l+r)/2;

  if (og<=mid)

  {

  rx[y]=rx[x]; 

  updata(l,mid,lx[x],lx[y],og);
 }else

        {

        lx[y]=lx[x];

        updata(mid+1,r,rx[x],rx[y],og);
}

 }

void query(int l,int r,int x,int y,int sug)

 {

  if (l==r)

  {

  printf("%d\n",cow[l]);

  return;
 }
int mid=(l+r)/2;
if (sum[lx[y]]-sum[lx[x]]<sug)
{
query(mid+1,r,rx[x],rx[y],sug-(sum[lx[y]]-sum[lx[x]]));
}else
 {
  query(l,mid,lx[x],lx[y],sug);
 }

 }

int main()

{
n=read();
for (int i=1;i<=n;++i)
{
  a[i].zhi=read(); a[i].q=i;

}
cnt=0;
sort(a+1,a+n+1,cmp);
for (int i=1;i<=n;++i)
 {
  ++cnt; b[a[i].q]=cnt; 
  cow[cnt]=a[i].q;
 }
for (int i=1;i<=n-1;++i)
{
  int u,v; u=read(); v=read(); insert(u,v); insert(v,u);
}
tail=0;
dfs(1,0);

    int q=read();

    opp=0;

    for (int i=1;i<=tail;++i)

     {

      updata(1,cnt,dui[i-1],dui[i],b[qg[i]]);
}

    for (int i=1;i<=q;++i)

     {

        int x,k;

        x=read(); k=read()*2;

        query(1,cnt,dui[st[x]-1],dui[en[x]],k);
}

}