bzoj 1803（主席树）

1803: Spoj1487 Query on a tree III

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 419  Solved: 184
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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：首先dfs求出dfs序列，然后就将树的数据化到序列上了，然后在序列上找K大，明显主席树。然而输出有点奇怪，输出的是值所对应得点，那只要多储存一个值，

就是到第i棵线段树，保存更改的最小面的那个值所对应的最后面的对应得点。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define rep(i,j,k) for(int i = j; i <= k; i++)#define MAX 200009#define u 18using namespace std;int to[1 * MAX], next[1 * MAX], head[1 * MAX], a[MAX], c[MAX];int n, m, k, dfn[MAX * 1], tree[u * MAX], child[u * MAX][2], size[u * MAX];int tot = 0, DfsClock = 0, sum = 0, first[MAX * 1], last[1 * MAX], num = 0;inline void add (int x, int y){  to[++tot] = y;  next[tot] = head[x];  head[x] = tot;}void dfs (int x, int fa){  dfn[first[x] = ++DfsClock] = x;  for (int i = head[x]; i; i = next[i])    if (to[i] != fa)      dfs (to[i], x);  dfn[last[x] = ++DfsClock] = x;}inline int add (int l, int r, int pos, int root){  int New = ++sum, mid = (l + r) >> 1;  size[New] = size[root] + 1;  if (l == r)    return New;  if (pos <= mid)    child[New][0] = add (l, mid, pos, child[root][0]), child[New][1] = child[root][1];  else    child[New][0] = child[root][0], child[New][1] = add (mid + 1, r, pos, child[root][1]);  return New;}struct wbysr{  int value, id;}e[MAX];bool cmp (wbysr a1, wbysr a2){  return a1.value < a2.value;}int main(){  scanf ("%d", &n);  rep (i, 1, n)    scanf ("%d", &a[i]), e[i].value = a[i], e[i].id = i;  sort (e + 1, e + 1 + n, cmp);  e[0].value = e[1].value - 90;  rep (i, 1, n)    if (e[i].value != e[i-1].value)      c[++num] = e[i].value;  rep (i, 1, n - 1)  {    int a1, a2;    scanf ("%d%d", &a1, &a2);    add (a1, a2);    add (a2, a1);  }  dfs (1,0);  tree[0] = 0;  size[0] = 0;  rep (i, 1, DfsClock)    tree[i] = add (1, num, lower_bound (c + 1, c + 1 + num, a[dfn[i]]) - c, tree[i - 1]);  scanf ("%d", &m);  while (m--)  {    int x;    scanf ("%d%d", &x, &k);    k *= 2;    int t0 = tree[first[x] - 1], t1 = tree[last[x]];    int l = 1, r = num;    while (l <= r)    {      int now = size[child[t1][0]] - size[child[t0][0]];      int mid = (l + r) >> 1;      if (k <= now)        t0 = child[t0][0], t1 = child[t1][0], r = mid;      else        t0 = child[t0][1], t1 = child[t1][1], k -= now, l = mid + 1;    }    printf ("%d\n", e[r].id);  }  return 0;}