misc
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Benjamin Qi
parent
486b06a350
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178308bbd5
5 changed files with 84 additions and 19 deletions
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@ -168,6 +168,65 @@ Although both Python and Java receive two times the C++ time limit in USACO, thi
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</details>
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<details>
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<summary>Comparable C++ Solution (< 700ms)</summary>
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```cpp
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#include <bits/stdc++.h>
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using namespace std;
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typedef vector<int> vi;
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const int MX = 1e5+5;
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int loc[MX], comp[MX], lhs[MX], rhs[MX], wei[MX];
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vi adj[MX];
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int n,m;
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void dfs(int cur, int label) {
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if (comp[cur] == label) return;
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comp[cur] = label;
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for (int c: adj[cur]) dfs(c,label);
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}
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bool valid(int minW) {
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for (int i = 0; i < n; ++i) {
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comp[i] = -1;
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adj[i].clear();
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}
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for (int i = 0; i < m; ++i) if (wei[i] >= minW)
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adj[lhs[i]].push_back(rhs[i]), adj[rhs[i]].push_back(lhs[i]);
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int numComps = 0;
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for (int i = 0; i < n; ++i) if (comp[i] < 0)
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dfs(i,numComps++);
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for (int i = 0; i < n; ++i)
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if (comp[i] != comp[loc[i]]) return 0;
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return 1;
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}
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int main() {
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ios_base::sync_with_stdio(0); cin.tie(0);
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freopen("wormsort.in","r",stdin);
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freopen("wormsort.out","w",stdout);
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cin >> n >> m;
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for (int i = 0; i < n; ++i) cin >> loc[i], loc[i] --;
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for (int i = 0; i < m; ++i) {
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cin >> lhs[i], lhs[i] --;
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cin >> rhs[i], rhs[i] --;
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cin >> wei[i];
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}
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int minW = 0, maxW = (int)1e9+1;
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while (minW != maxW) {
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int mid = (minW+maxW+1)/2;
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if (valid(mid)) minW = mid;
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else maxW = mid-1;
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}
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if (minW > 1e9) minW = -1;
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cout << minW;
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}
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```
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</details>
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## Other
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- Python lacks a data structure that keeps its keys in sorted order (the equivalent of `set` in C++), which is required for some silver problems.
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@ -5,6 +5,10 @@ author: Benjamin Qi
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order: 7
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---
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Shortening code and making it more (un?)readable.
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<!-- END DESCRIPTION -->
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## Introduction
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- [GeeksForGeeks](https://www.geeksforgeeks.org/cc-preprocessors/#:~:text=Macros%3A%20Macros%20are%20a%20piece,used%20to%20define%20a%20macro.)
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@ -1,7 +1,7 @@
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---
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slug: /silver/complexity
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title: "Time & Space Complexity"
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author: Darren Yao
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title: "Time Complexity"
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author: Darren Yao, Benjamin Qi
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order: 0
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---
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@ -113,12 +113,13 @@ Complexity factors that come from some common algorithms and data structures are
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- Mathematical formulas that just calculate an answer: $O(1)$
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- Unordered set/map: $O(1)$ per operation
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- Binary search: $O(\log n)$
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- Ordered set/map or priority queue: $O(log n)$ per operation
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- Ordered set/map or priority queue: $O(\log n)$ per operation
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- Prime factorization of an integer, or checking primality or compositeness of an integer naively: $O(\sqrt{n})$
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- Reading in $n$ items of input: $O(n)$
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- Iterating through an array or a list of $n$ elements: $O(n)$
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- Sorting: usually $O(n \log n)$ for default sorting algorithms (mergesort, for example Collections.sort or Arrays.sort on objects)
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- Java Quicksort Arrays.sort function on primitives on pathological worst-case data sets, don't use this in CodeForces rounds
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- Sorting: usually $O(n \log n)$ for default sorting algorithms (mergesort, for example `Collections.sort` or `Arrays.sort` on objects)
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- Java Quicksort `Arrays.sort` function on primitives: $O(n^2)$
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- on pathological worst-case data sets, don't use this in CodeForces rounds
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- Iterating through all subsets of size $k$ of the input elements: $O(n^k)$. For example, iterating through all triplets is $O(n^3)$.
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- Iterating through all subsets: $O(2^n)$
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- Iterating through all permutations: $O(n!)$
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@ -126,16 +127,17 @@ Complexity factors that come from some common algorithms and data structures are
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Here are conservative upper bounds on the value of $n$ for each time complexity. You can probably get away with more than this, but this should allow you to quickly check whether an algorithm is viable.
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- $n$ | Possible complexities
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- $n \le 10$ | $O(n!)$, $O(n^7)$, $O(n^6)$
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- $n \le 20$ | $O(2^n \cdot n)$, $O(n^5)$
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- $n \le 80$ | $O(n^4)$
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- $n \le 400$ | $O(n^3)$
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- $n \le 7500$ | $O(n^2)$
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- $n \le 7 \cdot 10^4$ | $O(n \sqrt n)$
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- $n \le 5 \cdot 10^5$ | $O(n \log n)$
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- $n \le 5 \cdot 10^6$ | $O(n)$
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- $n \le 10^{18}$ | $O(\log^2 n)$, $O(\log n)$, $O(1)$
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| $n$ | Possible complexities |
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| --------------------- | ----------------------------------- |
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| $n \le 10$ | $O(n!)$, $O(n^7)$, $O(n^6)$ |
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| $n \le 20$ | $O(2^n \cdot n)$, $O(n^5)$ |
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| $n \le 80$ | $O(n^4)$ |
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| $n \le 400$ | $O(n^3)$ |
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| $n \le 7500$ | $O(n^2)$ |
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| $n \le 7 \cdot 10^4$ | $O(n \sqrt n)$ |
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| $n \le 5 \cdot 10^5$ | $O(n \log n)$ |
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| $n \le 5 \cdot 10^6$ | $O(n)$ |
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| $n \le 10^{18}$ | $O(\log^2 n)$, $O(\log n)$, $O(1)$ |
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## Other Resources
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@ -8,7 +8,7 @@ prerequisites:
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- Silver - Sorting
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---
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**Two pointers** refers to iterating two monotonic pointers across an array to search for a pair of indices satisfying some condition in O(n).
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**Two pointers** refers to iterating two monotonic pointers across an array to search for a pair of indices satisfying some condition in $O(n)$ time.
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<!-- END DESCRIPTION -->
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@ -27,11 +27,11 @@ prerequisites:
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- [Cellular Network](http://codeforces.com/problemset/problem/702/C) [](48)
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- [USB vs. PS/2](http://codeforces.com/problemset/problem/762/B) [](53)
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- [K-Good Segment](http://codeforces.com/problemset/problem/616/D) [](53)
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- [(Long Title))](http://codeforces.com/problemset/problem/814/C) [](54)
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- [(Long Title)](http://codeforces.com/problemset/problem/814/C) [](54)
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- [Jury Meeting](http://codeforces.com/problemset/problem/853/B) [](90)
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- USACO
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- [Gold Haybale Feast](http://usaco.org/index.php?page=viewproblem2&cpid=767)
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## Extensions
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The two pointers technique can be extended to the two dimensional sliding window algorithm. If you're looking for a more challenging problem using 2-D sliding window, see [USACO Plat Fort Moo](http://usaco.org/index.php?page=viewproblem2&cpid=600)
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The two pointers technique can be extended to the two dimensional sliding window algorithm. If you're looking for a more challenging problem using 2-D sliding window, see [USACO Plat Fort Moo](http://usaco.org/index.php?page=viewproblem2&cpid=600).
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@ -5,7 +5,7 @@ author: Eric Wei (incomplete)
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order: 6
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---
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Given an array of size $N$, answer $Q$ queries of the following form: Find the sum of all elements between indices $i$ and $j$.
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> Given an array of size $N$, answer $Q$ queries of the following form: Find the sum of all elements between indices $i$ and $j$.
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<!-- END DESCRIPTION -->
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