目录

一、题目

1、题目描述

2、输入输出

2.1输入

2.2输出

3、原题链接

二、解题报告

1、思路分析

2、复杂度

3、代码详解

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一、题目

1、题目描述

2、输入输出

2.1输入

2.2输出

3、原题链接

F - Rated Range

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二、解题报告

1、思路分析

考虑定义 f(i, j) 为 初始分数为j，经过i 场比赛后的最终分数

f(i, j) = f(i - 1, j) + 1，如果 l[i] <= f(i - 1, j) <= r[i]

否则，f(i, j) = f(i - 1, j)

显然具有单调性 f(i, j + 1) >= f(i, j)

那么说明 f(i) 单调不降

那么我们用懒标记线段树维护 f(i)，每次读入 l(i) r(i)，对f 值 在 l(i) 和 r(i) 之间的线段进行 + 1即可

2、复杂度

时间复杂度： O(nlogn + q)空间复杂度：O(nlogn)

3、代码详解

 ​

#include <bits/stdc++.h>using i64 = long long;
using u64 = unsigned long long;
using u32 = unsigned;
using u128 = unsigned __int128;template<class Info, class Tag>
struct LazySegmentTree {int n;std::vector<Info> info;std::vector<Tag> tag;LazySegmentTree() : n(0) {}LazySegmentTree(int _n, Info _v = Info()) {init(_n, _v);}template<class T>LazySegmentTree(std::vector<T> _init) {init(_init);}void init(int _n, const Info &_v = Info()) {init(std::vector<Info>(_n, _v));}template<class T>void init(const std::vector<T> &_init) {n = _init.size();info.assign(4 << std::__lg(n), Info());tag.assign(4 << std::__lg(n), Tag());auto build = [&](auto &&self, int p, int l, int r) -> void {if (l == r) {info[p] = _init[l];return;}int m = (l + r) / 2;self(self, p * 2, l, m);self(self, p * 2 + 1, m + 1, r);pull(p);};build(build, 1, 0, n - 1);}void pull(int p) {info[p] = info[p * 2] + info[p * 2 + 1];}void apply(int p, const Tag &v) {info[p].apply(v);tag[p].apply(v);}void push(int p) {apply(2 * p, tag[p]);apply(2 * p + 1, tag[p]);tag[p] = Tag{};}void modify(int p, int l, int r, int x, const Info &v) {if (l == r) {info[p] = v;return;}int m = (l + r) / 2;push(p);if (x < m)modify(p * 2, l, m, x, v);elsemodify(p * 2 + 1, m + 1, r, x, v);pull(p);}void modify(int x, const Info &v) {modify(1, 0, n - 1, x, v);}Info rangeQuery(int p, int l, int r, int x, int y) {if (l > y || r < x)return Info{};if (x <= l && r <= y) return info[p];int m = (l + r) / 2;push(p);auto t = rangeQuery(p * 2, l, m, x, y) + rangeQuery(p * 2 + 1, m + 1, r, x, y);return t;}Info rangeQuery(int l, int r) {return rangeQuery(1, 0, n - 1, l, r);}void rangeApply(int p, int l, int r, int x, int y, const Tag &v) {if (l > y || r < x) return;if (x <= l && r <= y) {apply(p, v);return;}int m = (l + r) / 2;push(p);rangeApply(p * 2, l, m, x, y, v);rangeApply(p * 2 + 1, m + 1, r, x, y, v);pull(p);}void rangeApply(int l, int r, const Tag &v) {rangeApply(1, 0, n - 1, l, r, v);}template<class F>int findFirst(int p, int l, int r, int x, int y, F pred) {if (l > y || r < x || !pred(info[p])) {return -1;}if (l == r)return l;int m = (l + r) / 2;push(p);int res = findFirst(p * 2, l, m, x, y, pred);if (res == -1)res = findFirst(p * 2 + 1, m + 1, r, x, y, pred);return res;}template<class F>int findFirst(int l, int r, F pred) {return findFirst(1, 0, n - 1, l, r, pred);}template<class F>int findLast(int p, int l, int r, int x, int y, F pred) {if (l > y || r < x || !pred(info[p])) {return -1;}if (l == r)return l;int m = (l + r) / 2;push(p);int res = findLast(p * 2 + 1, m + 1, r, x, y, pred);if (res == -1)res = findLast(p * 2, l, m, x, y, pred);return res;}template<class F>int findLast(int l, int r, F pred) {return findLast(1, 0, n - 1, l, r, pred);}
};struct Tag{int x = 0;void apply(const Tag &t) {if (t.x) {x += t.x;}}
};struct Info{int max = 0;void apply(const Tag &t) {max += t.x;}friend Info operator+ (const Info &a, const Info &b) {return {std::max(a.max, b.max)};}
};struct F0{int t = 0;bool operator()(const Info& v) {return v.max >= t;}
};struct F1{int t = 0;bool operator()(const Info& v) {return v.max > t;}
};int main()
{std::ios::sync_with_stdio(false);std::cin.tie(nullptr);int n;std::cin >> n;constexpr int N = 5E5;std::vector<Info> init(N + 1);for (int i = 0; i <= N; ++ i) {init[i].max = i;}LazySegmentTree<Info, Tag> sgt(init);for (int i = 0; i < n; ++ i) {int l, r;std::cin >> l >> r;int L = sgt.findFirst(1, N, F0{l});int R = sgt.findFirst(1, N, F1{r});if (R == -1) {R = N + 1;}sgt.rangeApply(L, R - 1, {1});}int q;std::cin >> q;while (q --) {int x;std::cin >> x;std::cout << sgt.rangeQuery(x, x).max << '\n';}return 0;
}