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adj-diff-tree use monoid; add slide-window;
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,34 @@ | ||
| /* | ||
| 最后修改: | ||
| 20241013 | ||
| 测试环境: | ||
| gcc11.2,c++14 | ||
| clang12.0,C++14 | ||
| msvc14.2,C++14 | ||
| */ | ||
| #ifndef __OY_SLIDEWINDOW__ | ||
| #define __OY_SLIDEWINDOW__ | ||
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| #include <algorithm> | ||
| #include <cstdint> | ||
| #include <numeric> | ||
| #include <vector> | ||
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| namespace OY { | ||
| namespace WINDOW { | ||
| using size_type = uint32_t; | ||
| template <typename Callback, typename ShortenLeft, typename LengthenRight> | ||
| void solve(size_type length, size_type window_len, Callback &&call, ShortenLeft &&left_call, LengthenRight &&right_call) { | ||
| size_type l = 0, r = 0; | ||
| while (r != window_len) right_call(r++); | ||
| call(l, r - 1); | ||
| while (r != length) { | ||
| left_call(l++); | ||
| right_call(r++); | ||
| call(l, r - 1); | ||
| } | ||
| } | ||
| } | ||
| } | ||
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| #endif |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,63 @@ | ||
| ### 一、模板类别 | ||
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| 数据结构：滑动窗口。 | ||
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| 练习题目： | ||
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| 1. [3321. 计算子数组的 x-sum II](https://leetcode.cn/problems/find-x-sum-of-all-k-long-subarrays-ii) | ||
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| ### 二、模板功能 | ||
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| 本模板提供一个滑动窗口模板。 | ||
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| 滑动窗口问题中，需要知道区间总长度、窗口长度、窗口左端收缩的回调函数、窗口右端收缩的回调函数、窗口就绪的回调函数。只要传递这些参数，本模板即可自动运行。 | ||
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| ### 三、模板示例 | ||
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| ```c++ | ||
| #include "IO/FastIO.h" | ||
| #include "MISC/SlideWindow.h" | ||
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| #include <map> | ||
| void test() { | ||
| // 窗口数颜色 | ||
| cout << "count color count:\n"; | ||
| int a[] = {4, 7, 5, 1, 8, 1, 4, 4, 1}; | ||
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| int n = 9; | ||
| int window_len = 4; | ||
| std::map<int, int> mp; | ||
| auto call = [&](int l, int r) { | ||
| cout << "color count of ["; | ||
| for (int i = l; i <= r; i++) cout << a[i] << " ]"[i == r]; | ||
| cout << " = " << mp.size() << endl; | ||
| }; | ||
| auto left_call = [&](int i) { | ||
| auto &cnt = mp[a[i]]; | ||
| if (!--cnt) mp.erase(a[i]); | ||
| }; | ||
| auto right_call = [&](int i) { | ||
| auto &cnt = mp[a[i]]; | ||
| ++cnt; | ||
| }; | ||
| OY::WINDOW::solve(n, window_len, call, left_call, right_call); | ||
| } | ||
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| int main() { | ||
| test(); | ||
| } | ||
| ``` | ||
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| ``` | ||
| #输出如下 | ||
| count color count: | ||
| color count of [4 7 5 1] = 4 | ||
| color count of [7 5 1 8] = 4 | ||
| color count of [5 1 8 1] = 3 | ||
| color count of [1 8 1 4] = 3 | ||
| color count of [8 1 4 4] = 3 | ||
| color count of [1 4 4 1] = 2 | ||
| ``` | ||
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|
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,42 @@ | ||
| #include "IO/FastIO.h" | ||
| #include "MISC/SlideWindow.h" | ||
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| #include <map> | ||
| void test() { | ||
| // 窗口数颜色 | ||
| cout << "count color count:\n"; | ||
| int a[] = {4, 7, 5, 1, 8, 1, 4, 4, 1}; | ||
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| int n = 9; | ||
| int window_len = 4; | ||
| std::map<int, int> mp; | ||
| auto call = [&](int l, int r) { | ||
| cout << "color count of ["; | ||
| for (int i = l; i <= r; i++) cout << a[i] << " ]"[i == r]; | ||
| cout << " = " << mp.size() << endl; | ||
| }; | ||
| auto left_call = [&](int i) { | ||
| auto &cnt = mp[a[i]]; | ||
| if (!--cnt) mp.erase(a[i]); | ||
| }; | ||
| auto right_call = [&](int i) { | ||
| auto &cnt = mp[a[i]]; | ||
| ++cnt; | ||
| }; | ||
| OY::WINDOW::solve(n, window_len, call, left_call, right_call); | ||
| } | ||
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| int main() { | ||
| test(); | ||
| } | ||
| /* | ||
| #输出如下 | ||
| count color count: | ||
| color count of [4 7 5 1] = 4 | ||
| color count of [7 5 1 8] = 4 | ||
| color count of [5 1 8 1] = 3 | ||
| color count of [1 8 1 4] = 3 | ||
| color count of [8 1 4 4] = 3 | ||
| color count of [1 4 4 1] = 2 | ||
| */ |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,67 @@ | ||
| #include "DS/MonoAVL.h" | ||
| #include "DS/MonoSplay.h" | ||
| #include "IO/LeetcodeIO.h" | ||
| #include "MISC/SlideWindow.h" | ||
| using namespace std; | ||
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| /* | ||
| [3321. 计算子数组的 x-sum II](https://leetcode.cn/problems/find-x-sum-of-all-k-long-subarrays-ii) | ||
| */ | ||
| /** | ||
| * 滑动窗口模板，平衡树模板 | ||
| */ | ||
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| class Solution { | ||
| public: | ||
| vector<long long> findXSum(vector<int> &nums, int k, int x) { | ||
| // 本平衡树维护两个属性的和 | ||
| // 属性一是频率 | ||
| // 属性二是键和频率的乘积 | ||
| struct Monoid { | ||
| using value_type = pair<int, long long>; | ||
| static value_type op(value_type x, value_type y) { return {x.first + y.first, x.second + y.second}; } | ||
| static value_type identity() { return {0, 0}; } | ||
| }; | ||
| using Tree = OY::MONOAVL::Tree<Monoid, false>; | ||
| // using Tree = OY::MONOSPLAY::Tree<Monoid, false>; | ||
| Tree S; | ||
| // mp 维护频率 | ||
| unordered_map<int, int> mp; | ||
| vector<long long> ans; | ||
| // 给固定大小的滑动窗口传递回调函数 | ||
| // 对已就绪的窗口调用的回调 | ||
| auto call = [&](int l, int r) { | ||
| if (S.size() < x) | ||
| ans.push_back(S.query_all().second); | ||
| else | ||
| ans.push_back(S.query(S.size() - x, S.size() - 1).second); | ||
| }; | ||
| // 左端点右移的回调 | ||
| auto left_call = [&](int i) { | ||
| auto &cnt = mp[nums[i]]; | ||
| if (cnt) S.erase_by_comparator({cnt, 1ll * nums[i] * cnt}); | ||
| --cnt; | ||
| if (cnt) S.insert_by_comparator({cnt, 1ll * nums[i] * cnt}); | ||
| }; | ||
| // 右端点右移的回调 | ||
| auto right_call = [&](int i) { | ||
| auto &cnt = mp[nums[i]]; | ||
| if (cnt) S.erase_by_comparator({cnt, 1ll * nums[i] * cnt}); | ||
| ++cnt; | ||
| if (cnt) S.insert_by_comparator({cnt, 1ll * nums[i] * cnt}); | ||
| }; | ||
| OY::WINDOW::solve(nums.size(), k, call, left_call, right_call); | ||
| return ans; | ||
| } | ||
| }; | ||
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| #ifdef OY_LOCAL | ||
| int main() { | ||
| REGISTER_CONSTRUCTOR_SOLUTION; | ||
| REGISTER_MEMBERFUNCTION_SOLUTION(findXSum); | ||
| while (true) { | ||
| executor.constructSolution(); | ||
| executor.executeSolution(); | ||
| } | ||
| } | ||
| #endif |
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