bzoj 1803(主席树+dfs序)

最新推荐文章于 2021-09-14 17:59:37 发布

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本文介绍了一道经典的树形结构数据处理问题——查询节点子树中第k大的标签值。通过使用主席树和DFS序的方法，实现了高效的查询算法。

1803: Spoj1487 Query on a tree III

Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 510 Solved: 221
[ Submit][ Status][ Discuss]

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

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);
}
}