主席树区间第K大

主席树的实质其实还是一颗线段树， 然后每一次修改都通过上一次的线段树，来添加新边，使得每次改变就改变logn个节点，很多节点重复利用，达到节省空间的目的。

1.不带修改的区间第K大。

HDU-2665 模板题

代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb push_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int INF = 0x3f3f3f3f;
 const LL mod = 1e9+;
 const int N = 1e5+;
 struct Node{
     int l, r;
     int num;
 }tree[N*];
 int tot = ;
 int root[N];
 int Rank[N];
 int b[N];
 struct N{
     int a;
     int id;
     bool operator < (const N & x) const{
         return a < x.a;
     }
 }A[N];

 int Build(int l, int r){
     int now = ++tot;
     tree[now].num = ;
     if(l == r) return now;
     int m = l+r >> ;
     tree[now].l = Build(l,m);
     tree[now].r = Build(m+,r);
     return now;
 }
 int update(int pre, int l, int r, int k){
     int now = ++tot;
     tree[now].num = ;
     tree[now].l = tree[pre].l;
     tree[now].r = tree[pre].r;
     if(l < r) {
         int m = l+r >> ;
         if(k <= m) tree[now].l = update(tree[pre].l ,l, m, k);
         else tree[now].r = update(tree[pre].r, m+, r, k);
         tree[now].num = tree[tree[now].l].num + tree[tree[now].r].num;
     }
     else tree[now].num = ;
     return now;
 }
 int query(int u, int v, int l, int r,int k){
     if(l == r) return l;
     int m = l+r >> ;
     int num1 = tree[tree[v].l].num - tree[tree[u].l].num;
     if(num1 >= k) return query(tree[u].l, tree[v].l, l, m, k);
     return query(tree[u].r, tree[v].r, m+,r,k-num1);
 }
 int main(){
     ///Fopen;
     int t, n, m;
     scanf("%d", &t);
     while(t--){
         scanf("%d%d", &n, &m);
         for(int i = ; i <= n; i++){
             scanf("%d", &A[i].a);
             A[i].id = i;
         }
         sort(A+, A++n); /// 离散化
         int t = ;
         A[].a = -;
         for(int i = ; i <= n; i++){
             if(A[i].a != A[i-].a) t++, Rank[t] = A[i].a;
             b[A[i].id] = t;
         }
         tot = ;
         root[] = Build(,t);
         for(int i = ; i <= n; i++)
             root[i] = update(root[i-], , t, b[i]);
         while(m--){
             int l, r, k;
             scanf("%d%d%d", &l, &r, &k);
             printf("%d\n", Rank[query(root[l-], root[r],, t, k)]);
         }
     }
     return ;
 }
 /*
 1
 10 1
 1 4 2 3 5 6 7 8 9 0
 1 3 2

 */