主席树模板

用这个模板可以直接A掉 HDU 2665 Kth number 这题了！

/* 主席树求区间第K大模板：
* 模板特殊说明：
* 每棵树是维护从1开始到cnt的下标信息
*/
#include <stdio.h>
#include <algorithm>
#define maxn 100010

using namespace std;

int T, n, m, tot, a[maxn], b[maxn], cnt;
//T为输入cas组数，n为原数列数组个数，m为询问个数，tot为建树时新节点
//a为原数列，b为用来离散化的数列，其unique后元素个数为cnt

struct N { int ls, rs, w; } tree[30 * maxn];
//ls为左儿子节点下标，rs为右儿子节点下标，w为此区间内元素数量
int roots[maxn];
//roots记录每棵树的根节点下标

int build_tree(int l, int r) { //先建造一棵空树
int newnode = tot++;
tree[newnode].w = 0;
if (l != r) {
int mid = (l + r) / 2;
tree[newnode].ls = build_tree(l, mid);
tree[newnode].rs = build_tree(mid + 1, r);
}
return newnode;
}

int updata(int rt, int pos, int val) { //rt为根节点，在pos位置上加val的值
int newnode = tot++, tmp = newnode;
tree[newnode].w = tree[rt].w + val;
int l = 1, r = cnt;
while (l < r) {
int mid = (l + r) / 2;
if (pos <= mid) {
tree[newnode].ls = tot++;
tree[newnode].rs = tree[rt].rs;
newnode = tree[newnode].ls;
rt = tree[rt].ls;
r = mid;
}
else {
tree[newnode].ls = tree[rt].ls;
tree[newnode].rs = tot++;
newnode = tree[newnode].rs;
rt = tree[rt].rs;
l = mid + 1;
}
tree[newnode].w = tree[rt].w + val;
}
return tmp;
}

int query(int rt1, int rt2, int k) { //询问根节点分别为rt1，rt2两棵树之间第k大的值
int l = 1, r = cnt;
while (l < r) {
int mid = (l + r) / 2;
int tmp = tree[tree[rt2].ls].w - tree[tree[rt1].ls].w;
if (tmp >= k) {
rt1 = tree[rt1].ls;
rt2 = tree[rt2].ls;
r = mid;
}
else {
k -= tmp;
rt1 = tree[rt1].rs;
rt2 = tree[rt2].rs;
l = mid + 1;
}
}
return l;
}

int main() {
//freopen("in.in", "r", stdin);
//freopen("out.out", "w", stdout);
scanf("%d", &T);
while (T--) { //读入数据组数
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
b[i - 1] = a[i];
}
sort(b, b + n);
cnt = unique(b, b + n) - b; //离散化
tot = 0;
roots[0] = build_tree(1, cnt);
for (int i = 1; i <= n; i++) {
int tmp = (int)(lower_bound(b, b + cnt, a[i]) - b) + 1;
roots[i] = updata(roots[i - 1], tmp, 1); //在上一次的基础上建树
}
int l,r,k;
for(int i=0;i<m;i++){
scanf("%d %d %d",&l,&r,&k); //读入求l和r区间第k大的数
int tmp = query(roots[l-1],roots[r],k);
printf("%d\n",b[tmp - 1]);
}
}
return 0;
}