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P3834 【模板】可持久化线段树 2
主席树
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#include <algorithm>
#include <cstdio>
#include <iostream>
const int N = 2e5 + 3;
int n, m, a[N], p[N], T[N], s, cnt;//T树根编号
int sum[N << 5], L[N << 5], R[N << 5];//L左孩子编号,sum包含元素个数
int build(int l, int r) {
int rt = ++cnt;
if (l < r) {
int mid = (l + r) >> 1;
L[rt] = build(l, mid);
R[rt] = build(mid + 1, r);
}
return rt;
}//build一棵空树
int update(int pre, int l, int r, int x) {
int rt = ++cnt;//新建节点
L[rt] = L[pre], R[rt] = R[pre], sum[rt] = sum[pre] + 1;//左右儿子继承之前版本,元素个数在前版本基础上+1
if (l < r) {
int mid = (l + r) >> 1;
if (x <= mid) L[rt] = update(L[pre], l, mid, x);
else R[rt] = update(R[pre], mid + 1, r, x);
}
return rt;
}
int query(int u, int v, int l, int r, int k) {
if (l >= r) return l;
int x = sum[L[v]] - sum[L[u]], mid = (l + r) >> 1;
if (x >= k) return query(L[u], L[v], l, mid, k);
else return query(R[u], R[v], mid + 1, r, k - x);
}
int main() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
p[i] = a[i];
}
std::sort(p + 1, p + n + 1);
s = std::unique(p + 1, p + n + 1) - p - 1;
T[0] = build(1, s);
for (int i = 1; i <= n; ++i) {
T[i] = update(T[i - 1], 1, s, std::lower_bound(p + 1, p + 1 + s, a[i]) - p);
}
for (int i = 1, l, r, k; i <= m; ++i) {
scanf("%d %d %d", &l, &r, &k);
printf("%d/n", p[query(T[l - 1], T[r], 1, s, k)]);
}
return 0;
}
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/281757.html