SPOJ 1487. Query on a tree III
SPOJ Problem Set (classical)SPOJ 1487. Query on a tree IIIProblem code: PT07J |
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 <= 105). 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 <= 104) which denotes the number of the queries. Each line of the nextm 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.
Example
Input: 5 1 3 5 2 7 1 2 2 3 1 4 3 5 4 2 3 4 1 3 2 3 2 Output: 5 4 5 5
---------------------------------------------------------------
题目大意：询问树上某棵子树上第K大的节点。
解题思路：用一次dfs把树搞成括号序列，记录每个节点的开始和结束时间戳，然后就变成了区间第k大值，用划分树模板即可。
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#define MID(x,y) ( ( x + y ) >> 1 )
#define L(x) ( x << 1 )
#define R(x) ( x << 1 | 1 )
#define BUG puts("here!!!")
#define LL long long
using namespace std;
const int N=100005;
int n,m;
int val[40][N],sorted[N],toleft[40][N];
bool cmp(int a,int b)
{
return a<b;
}
struct node
{
int l,r;
int getmid()
{
return (l+r)>>1;
}
} tree[N * 4];
void build(int rt,int l,int r,int d)
{
tree[rt].l=l;
tree[rt].r=r;
if(l==r) return;
int mid=(l+r) >> 1;
int same=mid-l+1;
for(int i=l; i<=r; ++i)
if (val[d][i]<sorted[mid])same--;
int lpos=l;
int rpos=mid+1;
for(int i=l; i<=r; ++i)
{
if(i==l)toleft[d][i]=0;
else toleft[d][i]=toleft[d][i - 1];
if(val[d][i]<sorted[mid])
{
toleft[d][i]++;
val[d+1][lpos++]=val[d][i];
}
else if(val[d][i]>sorted[mid])val[d+1][rpos++]=val[d][i];
else if(same)
{
same--;
toleft[d][i]++;
val[d+1][lpos++]=val[d][i];
}
else val[d+1][rpos++]=val[d][i];
}
build(L(rt),l,mid,d+1);
build(R(rt),mid+1,r,d+1);
}
int query(int rt, int l, int r, int d, int k)
{
if(l==r)return val[d][l];
int s,ss;
if(l==tree[rt].l)
{
s=toleft[d][r];
ss=0;
}
else
{
s=toleft[d][r]-toleft[d][l-1];
ss=toleft[d][l-1];
}
if(k<=s)
{
int left=tree[rt].l+ss;
int right=left+s-1;
return query(L(rt),left,right,d+1,k);
}
else
{
int b=l-tree[rt].l-ss;
int left=tree[rt].getmid()+b+1;
int right=left+r-l-s;
return query(R(rt),left,right,d+1,k-s);
}
}
int lab[N],one[N],two[N],eid,num;
int head[N],ed[N<<1],nxt[N<<1];
map<int,int>ha;
void addedge(int s,int e)
{
ed[eid]=e;
nxt[eid]=head[s];
head[s]=eid++;
}
void dfs(int s,int f)
{
one[s]=num;
sorted[num]=lab[s];
val[0][num++]=lab[s];
for(int i=head[s]; ~i; i=nxt[i])
if(ed[i]!=f)dfs(ed[i],s);
two[s]=num-1;
}
int main()
{
// freopen("/home/axorb/in","r",stdin);
ha.clear();
scanf("%d",&n);
eid=0;
memset(head,-1,sizeof(head));
for(int i=1; i<=n; i++)
{
scanf("%d",&lab[i]);
ha[lab[i]]=i;
}
for(int i=1; i<n; i++)
{
int a,b;
scanf("%d%d",&a,&b);
addedge(a,b);
addedge(b,a);
}
num=0;
dfs(1,-1);
sort(sorted,sorted+n,cmp);
build(1,0,n-1,0);
int q;
scanf("%d",&q);
while(q--)
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",ha[query(1,one[a],two[a],0,b)]);
}
}