Problems
Started writing conversion in string suffix structs
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Siyong
parent
b125a0be0f
commit
c6d997668a
3 changed files with 117 additions and 8 deletions
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@ -9,12 +9,20 @@ description: "?"
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frequency: 1
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---
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<!--
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TODO:
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- Add more resources
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-->
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import { Problem } from "../models";
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export const metadata = {
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problems: {
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general: [
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new Problem("CF", "Ciel the Commander", "problemset/problem/321/C", "Easy", false, ["Centroid"], ""),
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new Problem("Plat", "New Barns", "817", "Normal", true, ["Centroid"], ""),
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new Problem("CF", "Sherlock's bet to Moriarty", "contest/776/problem/F", "Normal", false, ["Centroid"], ""),
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new Problem("CF", "Digit Tree", "contest/715/problem/C", "Normal", false, ["Centroid", "NT"], ""),
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new Problem("CF", "Double Tree", "contest/1140/problem/G", "Normal", false, ["Centroid", "DP"], ""),
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@ -84,9 +92,43 @@ void centroid(int n = 1)
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</CPPSection>
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<JavaSection />
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<JavaSection>
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<PySection />
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<!-- Modified from above. I can't guarauntee it compiles and functions as expected -->
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```java
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boolean[] r = new boolean[MN];//removed
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int[] s = new int[MN];//subtree size
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public int dfs(int n, int p)
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{
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s[n] = 1;
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for(int x : a[n])
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if(x != p && !r[x])
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s[n] += dfs(x, n);
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return s[n];
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}
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public int get_centroid(int n, int ms, int p)//n = node, ms = size of tree, p = parent
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{
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for(int x : a[n])
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if(x != p && !r[x])
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if(s[x]*2 > ms)
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return get_centroid(x, ms, n);
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return n;
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}
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public void centroid(int n)
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{
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int C = get_centroid(n, dfs(n, 0), 0);
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//do something
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r[C] = 1;
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for(int x : a[C])
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if(!r[x])
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centroid(x);
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}
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```
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</JavaSection>
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</LanguageSection>
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@ -23,6 +23,7 @@ export const metadata = {
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new Problem("CSES", "Elevator Rides", "1653", "Normal", false, ["Bitmasks"], ""),
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new Problem("ojuz", "IZhO Bank", "IZhO14_bank", "Normal", false, ["Bitmasks"], ""),
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new Problem("YS", "Max Indep Set", "maximum_independent_set", "Normal", false, ["Bitmasks", "Meet in Middle"], ""),
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new Problem("CF", "Wise Men", "contest/1326/problem/F2", "Very Hard", false, ["Bitmasks", "DP", "SOS"], "Solve the case where for each binary string s, a 1 means that the adjacent men know each other, and the 0 means nothing: they can know each other or not."),
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],
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broken: [
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new Problem("CF", "Guards in the Storehouse", "problemset/problem/845/F", "Normal", false, [], ""),
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@ -44,6 +45,7 @@ You can often use this to solve subtasks.
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<Resource source="CPH" title="10 (Bit Manipulation), 19.2 (Hamiltonian Paths)"> </Resource>
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<Resource source="CF" title="DP Over Subsets" url="blog/entry/337"> </Resource>
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<Resource source="HE" title="DP and Bit Masking" url="https://www.hackerearth.com/practice/algorithms/dynamic-programming/bit-masking/tutorial/"> </Resource>
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<Resource source="CF" title="SOS Dynamic Programming" url="blog/entry/45223"> Sum over Subsets DP </Resource>
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</Resources>
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## Problems
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@ -58,4 +60,4 @@ You can often use this to solve subtasks.
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(fill in? more probs?)
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<Problems problems={metadata.problems.broken} />
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<Problems problems={metadata.problems.broken} />
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@ -204,7 +204,7 @@ Naively, this would take up $O(N^2)$ memory, but *path compression* enables it t
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<Problems problems={metadata.problems.tree} />
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### Generate Suffix Array
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### Generate Suffix Array from Suffix Tree
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A suffix array can be generated by the suffix tree by taking the dfs traversal of the suffix tree.
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@ -217,12 +217,12 @@ A suffix array can be generated by the suffix tree by taking the dfs traversal o
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```cpp
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int N, sa[MN];//length of string, suffix array
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struct edge
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struct Edge
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{
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public:
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int n, l, r;//node, edge covers s[l..r]
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explicit operator bool() const {return n!=-1;}
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} c[MN*2][26];
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} c[MN*2][26]; // edges of a suffix tree
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void dfs(int n=0, int d=0)
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{
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@ -239,9 +239,74 @@ void dfs(int n=0, int d=0)
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```
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</CPPSection>
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<JavaSection />
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</LanguageSection>
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<PySection />
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### Generate Suffix Tree from Suffix Array
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Of course, the above operation can be reversed as well.
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Each element in the suffix array corresponds to a leaf in the suffix tree.
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The LCP array stores information about the Lowest Common Ancestor of two adjacent elements in the suffix array.
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Using these two pieces of information, we can construct the suffix tree from the suffix array in linear time.
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<LanguageSection>
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<CPPSection>
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```cpp
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//untested
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int N, sa[MN], lcp[MN];//length of string, suffix array, lcp array: lcp[i] stores longest common prefix of sa[i] and sa[i+1]
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struct Edge
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{
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public:
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int n, l, r; //node, edge covers s[l..r]
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explicit operator bool() const {return n!=-1;}
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} c[MN*2][26]; // edges of suffix tree
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void build_tree()
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{
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}
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```
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</CPPSection>
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</LanguageSection>
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### Generate Suffix Tree from Suffix Automaton
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One interesting thing about Suffix Trees and Suffix Automata is that the link tree of a Suffix Automaton is equivalent to the Suffix Tree of the reversed string.
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Since Suffix Automata are much easier to create than Suffix Trees, we can use this as an alternate method to build a Suffix Tree, all in linear time too!
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<LanguageSection>
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<CPPSection>
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<!-- https://codeforces.com/edu/course/2/lesson/2/2/practice/contest/269103/submission/85759835 - This submission contains both -->
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```cpp
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char s[MN]; //string
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int ord[MN]; // nodes representing prefixes of the string s
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int u[MN*2]; // whether the node has already been created
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int l[MN*2]; // link in suffix automaton
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Edge c[MN*2][27]; // edge of suffix tree (not automaton; structure of automaton is not necessary to build stree)
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void build_tree()
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{
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s[N] = 26; // terminator
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for(int i=N;i>=0;--i) ord[i]=append(ord[i+1], s[i]);
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for(int i=0,x,r,l;i<=N;++i)
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{
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x=ord[i], r=N+1;
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for(;x&&!u[x];x=l[x])
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{
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l=r-d[x]+d[l[x]];
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c[l[x]][s[l]]={x, l, r};
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r=l;
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u[x]=1;
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}
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}
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}
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```
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</CPPSection>
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</LanguageSection>
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