牛客小白月赛28 E 会当凌绝顶,一览众山小
会当凌绝顶,一览众山小
https://ac.nowcoder.com/acm/contest/7412/E
E 会当凌绝顶,一览众山小
题目地址:
基本思路:
没有什么思维难度,但是代码难度比较高,
做一个类似离散化的排序,然后就是建线段树,实现所有操作。
因为线段树就是天然二分结构,
所以实际上这里的所有操作都可以在线段树上实现,
这里算是给自己存一个模板。
参考代码:
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define int long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f
inline int read() {
int x = 0, neg = 1; char op = getchar();
while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
return neg * x;
}
inline void print(int x) {
if (x < 0) { putchar('-'); x = -x; }
if (x >= 10) print(x / 10);
putchar(x % 10 + '0');
}
const int maxn = (int)2e5 + 10;
#define ls (index << 1)
#define rs (index << 1 | 1)
struct SegmentTree {
struct Node {
int l, r, sum, mn, mx, id;
}tr[maxn * 4];
inline void push_up(int index) { // 带储存区间最小值对应下标的push_up;
tr[index].sum = tr[ls].sum + tr[rs].sum;
tr[index].mx = max(tr[ls].mx, tr[rs].mx);
if (tr[ls].mn <= tr[rs].mn) {
tr[index].mn = tr[ls].mn;
tr[index].id = tr[ls].id;
} else {
tr[index].mn = tr[rs].mn;
tr[index].id = tr[rs].id;
}
}
void build(int index, int l, int r, int *a) { // 建树;
tr[index].l = l, tr[index].r = r;
if (l == r) {
tr[index].id = l;
tr[index].sum = tr[index].mn = tr[index].mx = a[l];
return;
}
int mid = (l + r) >> 1;
build(ls, l, mid, a);
build(rs, mid + 1, r, a);
push_up(index);
}
void update(int index, int val, int x) { // 单点修改;
if (tr[index].l == tr[index].r) {
tr[index].sum = val;
tr[index].mn = val;
tr[index].mx = val;
return;
}
int mid = (tr[index].l + tr[index].r) >> 1;
if (x <= mid) update(ls, val, x);
else update(rs, val, x);
push_up(index);
}
int ask(int index, int x) { // 单点查询;
if (tr[index].l == tr[index].r) {
return tr[index].mn;
}
int mid = (tr[index].l + tr[index].r) >> 1;
if (x <= mid) return ask(ls, x);
else return ask(rs, x);
}
pii query_mn(int index, int l, int r) { // 查询区间最小值和它对应下标;
if (tr[index].l >= l && tr[index].r <= r) {
return mp(tr[index].mn, tr[index].id);
}
int mid = (tr[index].l + tr[index].r) >> 1;
if (r <= mid) return query_mn(ls, l, r);
else if (l > mid) return query_mn(rs, l, r);
else {
pii r1 = query_mn(ls, l, mid), r2 = query_mn(rs, mid + 1, r);
return r1.first <= r2.first ? r1 : r2;
}
}
int query_mx(int index, int l, int r) { // 查询区间最大值;
if (tr[index].l >= l && tr[index].r <= r) {
return tr[index].mx;
}
int mid = (tr[index].l + tr[index].r) >> 1;
if (r <= mid) return query_mx(ls, l, r);
else if (l > mid) return query_mx(rs, l, r);
else return max(query_mx(ls, l, mid), query_mx(rs, mid + 1, r));
}
int query_pos(int index, int l, int r, int val) { // 查询区间最后一个大于val的数;
if (tr[index].l == tr[index].r) {
if (tr[index].mx > val) return tr[index].l;
else return -1;
}
if (tr[index].l >= l && tr[index].r <= r) {
if (tr[index].mx <= val) return -1;
}
int mid = (tr[index].l + tr[index].r) >> 1;
if (r <= mid) return query_pos(ls, l, r, val);
else if (l > mid) return query_pos(rs, l, r, val);
// 如果是查询区间第一个大于val的数,就先往左边找;
int tmp = query_pos(rs, mid + 1, r, val);
if (tmp != -1) return tmp;
else return query_pos(ls, l, mid, val);
}
}smt;
int n,a[maxn];
struct Node{
int x,h,pos;
bool operator < (const Node &no) const{
return x < no.x;
}
}m[maxn];
int mm[maxn],ans[maxn];
signed main() {
IO;
cin >> n;
rep(i, 1, n) {
m[i].pos = i;
cin >> m[i].x >> m[i].h;
}
sort(m + 1, m + 1 + n);
rep(i, 1, n) a[i] = m[i].h, mm[m[i].pos] = i;
smt.build(1, 1, n, a);
rep(i, 1, n) {
int hig = smt.ask(1, mm[i]);
if (mm[i] != 1) {
int pos1 = smt.query_pos(1, 1, mm[i] - 1, hig);
if (pos1 != -1) smt.update(1, hig, pos1);
}
if (mm[i] != n) {
int mx = smt.query_mx(1, mm[i] + 1, n);
if (mx <= hig) {
int pos2 = smt.query_mn(1, mm[i] + 1, n).second;
smt.update(1, hig, pos2);
}
}
}
rep(i, 1, n) ans[m[i].pos] = smt.ask(1, i);
rep(i, 1, n) cout << ans[i] << ' ';
cout << '\n';
return 0;
}
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