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<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>reveal.js – The HTML Presentation Framework</title>
<meta name="description" content="A framework for easily creating beautiful presentations using HTML">
<meta name="author" content="Hakim El Hattab">
<meta name="apple-mobile-web-app-capable" content="yes">
<meta name="apple-mobile-web-app-status-bar-style" content="black-translucent">
<meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no, minimal-ui">
<link rel="stylesheet" href="css/reveal.css">
<link rel="stylesheet" href="css/theme/black.css" id="theme">
<!-- Code syntax highlighting -->
<link rel="stylesheet" href="lib/css/zenburn.css">
<!-- Printing and PDF exports -->
<script>
var link = document.createElement( 'link' );
link.rel = 'stylesheet';
link.type = 'text/css';
link.href = window.location.search.match( /print-pdf/gi ) ? 'css/print/pdf.css' : 'css/print/paper.css';
document.getElementsByTagName( 'head' )[0].appendChild( link );
</script>
<!--[if lt IE 9]>
<script src="lib/js/html5shiv.js"></script>
<![endif]-->
</head>
<body>
<div class="reveal">
<!-- Any section element inside of this container is displayed as a slide -->
<div class="slides">
<section>
<section>
<h1>Data Structure Basic</h1>
</section>
<section>
<h1>Who am I?</h1>
<ul>
<li>eddy1021,eddy,艾迪,欸底,...</li>
<li>PECaveros</li>
</ul>
</section>
<section>
<h2>What you will learn?</h2>
<ul>
<li>堆疊 stack</li>
<li>佇列 queue</li>
<li>雙端佇列 deque</li>
<li>優先佇列 priority queue</li>
</ul>
</section>
<section>
<h2>And also</h2>
<ul>
<li>並查集 disjoint set</li>
<li>線段樹 segment tree</li>
<li>樹堆 treap</li>
<li>...</li>
</ul>
</section>
<section>
<h1>Let's fight!</h1>
</section>
</section>
<!-- stack -->
<section>
<section>
<h1>堆疊 Stack</h1>
</section>
<section>
<h2>What is stack?</h2>
<ul>
<li>後進先出, Last In First Out</li>
<li>支援push, pop</li>
</ul>
</section>
<section>
<img src="images/stack.png">
</section>
<section>
<h2>How to do so?</h2>
<pre><code class="cpp" data-trim contenteditable>
#include <stack>
#include <stdio.h>
using namespace std;
int main(){
stack<int> s;
s.push( 514 ); s.push( 514514 );
printf( "%d\n" , (int)s.size() ); // 2
printf( "%d\n" , s.top() ); // 514514
s.pop();
printf( "%d\n" , s.top() ); // 514
s.pop();
printf( "%s\n" , s.empty() ? "empty" : "not empty" );
// empty
}
</code></pre>
</section>
<section>
<h1>常見題型</h1>
<ul>
<li>一臉堆疊樣</li>
<li>維護凸包</li>
<li>維護單調性</li>
</ul>
</section>
<section>
<p>寬 $N$ 的直方圖，第 $i$ 單位寬的部分高 $h_i$，問面積最大的矩形？</p>
<img src="images/poj2559.png">
</section>
<section>
<h2>枚舉左右界</h2>
<p>最佳矩形高度爲 $\min\{ h_L, h_{L+1}, \ldots, h_R \}$</p>
</section>
<section>
<h2>最佳左右界</h2>
<p>往外擴到底！</p>
<img src="images/poj2559-1.png">
</section>
<section>
<h2>單調性</h2>
<p>若存在 $j$ < $i$ 且 $h_j$ > $h_i$，則 $i$ 以後 $h_j$ 無法成爲區間極值</p>
</section>
<section>
<h2>計算最佳左右界</h2>
<ol>
<li>考慮 $h_i$</li>
<li>堆疊頂端 $h_j \ge h_i$，則pop </li>
<li>若堆疊爲空，最佳左(右)界爲1，否則爲 $j+1$(堆疊頂端爲 $h_j$)</li>
<li>push $h_i$</li>
</ol>
</section>
<section>
<h2>計算最佳左右界</h2>
<img src="images/poj2559-2.png">
</section>
<section>
<pre><code class="cpp" data-trim contenteditable>
stack<int> S;
for( int i = 1 ; i <= n ; i ++ ){
while( S.size() && h[ S.top() ] >= h[ i ] ) S.pop();
L[ i ] = 1 + ( S.size() ? S.top() : 0 );
S.push( i );
}
while( S.size() ) S.pop();
for( int i = n ; i >= 1 ; i -- ){
while( S.size() && h[ S.top() ] >= h[ i ] ) S.pop();
R[ i ] = ( S.size() ? S.top() : n + 1 ) - 1;
S.push( i );
}
</code></pre>
</section>
</section>
<!-- queue -->
<section>
<section>
<h1>佇列 Queue</h1>
</section>
<section>
<h2>What is queue?</h2>
<ul>
<li>先進先出, First In First Out</li>
<li>支援push, pop</li>
</ul>
</section>
<section>
<img src="images/queue.png">
</section>
<section>
<h2>How to do so?</h2>
<pre><code class="cpp" data-trim contenteditable>
#include <queue>
#include <stdio.h>
using namespace std;
int main(){
queue<int> q;
q.push( 514 ); q.push( 514514 );
printf( "%d\n" , (int)q.size() ); // 2
printf( "%d\n" , q.front() ); // 514
q.pop();
printf( "%d\n" , q.front() ); // 514514
q.pop();
printf( "%s\n" , q.empty() ? "empty" : "not empty" );
// empty
}
</code></pre>
</section>
<section>
<h1>常見題型</h1>
<ul>
<li>一臉佇列樣</li>
<li>Breadth First Search(BFS)!</li>
<li> <strike>中國人排隊問題</strike> </li>
</ul>
</section>
<section>
<p>$R \times C$的地圖上，有障礙物、人(恰一個)、起火點(若干個)，問人是否能在不受火災侵襲而逃出迷宮？</p>
<img src="images/uva11624.png">
</section>
<section>
<img src="images/uva11624-1.png">
</section>
<section>
<h2>火很多QQ</h2>
<p>要 BFS 很多次，複雜度 $O(R\times C \times R \times C)$</p>
</section>
<section>
<h1>全部火一起BFS!</h1>
<p>複雜度 $O(R\times C )$</p>
</section>
<section>
<pre><code class="cpp" data-trim contenteditable>
while( q.size() ){
PII tp = q.front(); q.pop();
int nr = tp.first;
int nc = tp.second;
for( int i = 0 ; i < 4 ; i ++ ){
int nxtr = nr + dr[ i ];
int nxtc = nc + dc[ i ];
if( !in( nxtr , nxtc ) )
ans = min( ans , dst[ nr ][ nc ] + 1 );
else if( m[ nxtr ][ nxtc ] != '#' &&
dst[ nr ][ nc ] + 1 < dst[ nxtr ][ nxtc ] ){
dst[ nxtr ][ nxtc ] = dst[ nr ][ nc ] + 1;
q.push( { nxtr , nxtc } );
}
}
}
</code></pre>
</section>
</section>
<!-- deque -->
<section>
<section>
<h1>雙端佇列 Deque</h1>
</section>
<section>
<h2>What is deque?</h2>
<ul>
<li>都進都出, Both In Both Out</li>
<li>支援兩端push, pop</li>
</ul>
</section>
<section>
<img src="images/deque.png">
</section>
<section>
<h2>How to do so?</h2>
<pre><code class="cpp" data-trim contenteditable>
#include <deque>
#include <stdio.h>
using namespace std;
int main(){
deque<int> dq;
dq.push_back( 514 ); // 514
dq.push_back( 514514 ); // 514 514514
dq.push_front( 514514514 ); // 514514514 514 514514
printf( "%d\n" , (int)dq.size() ); // 3
printf( "%d\n" , dq.front() ); // 514514514
printf( "%d\n" , dq.back() ); // 514514
dq.pop_front();
printf( "%d\n" , dq[ 1 ] ); // 514514
}
</code></pre>
<h3>STL 中 deque 支援random access!</h3>
</section>
<section>
<h1>常見題型</h1>
<ul>
<li>一臉得可樣</li>
<li>維護單調性與區間長度</li>
<li>DP斜率優化</li>
</ul>
</section>
<section>
<p>給長度 $N$ 的序列 $a$ 及正整數 $K$。求出所有 $b_i = max\{ a_{i-K+1}, a_{i-K+2}, \cdots, a_i \}$</p>
</section>
<section>
<p>當 i < j 且 $a_i$ < $a_j$，則 $a_i$ 就再也不可能是答案</p>
<p>同時維護長度！</p>
</section>
<section>
<pre><code class="cpp" data-trim contenteditable>
deque<int> dq;
for( int i = 1 ; i <= n ; i ++ ){
while( dq.size() && dq.front()<=i-k ) dq.pop_front();
while( dq.size() && a[dq.back()]<=a[i] ) dq.pop_back();
dq.push_back( i );
b[ i ] = a[ dq.front() ];
}
</code></pre>
</section>
</section>
<!-- priority queue -->
<section>
<section>
<h1>優先佇列</h1>
<h1>Priority Queue</h1>
</section>
<section>
<h2>What is priority queue?</h2>
<ul>
<li>VIP制</li>
<li>優先權越高，越早出來</li>
<li>支援push, pop</li>
</ul>
</section>
<section>
<img src="images/heap.png">
</section>
<section>
<h2>How to do so?</h2>
<pre><code class="cpp" data-trim contenteditable>
#include <queue>
#include <stdio.h>
using namespace std;
typedef priority_queue<int> Heap;
int main(){
Heap h;
h.push( 5 ); h.push( 1 ); h.push( 4 );
printf( "%d\n" , (int)h.size() ); // 3
printf( "%d\n" , h.top() ); // 5
h.pop();
printf( "%d\n" , h.top() ); // 4
h.push( 2 );
printf( "%d\n" , h.top() ); // 4
h.pop();
printf( "%d\n" , h.top() ); // 2
}
</code></pre>
<h3>STL 中優先權大者爲值較大者</h3>
</section>
<section>
<h2>自定義比較函式</h2>
<pre><code class="cpp" data-trim contenteditable>
struct cmp{
bool operator()( int a , int b ){
return a % 3 < b % 3;
}
};
int main(){
priority_queue<int, vector<int>, cmp> h;
h.push( 4 );
printf( "%d\n" , h.top() ); // 4, 4 % 3 == 1
h.push( 51 );
printf( "%d\n" , h.top() ); // 4, 51 % 3 == 0
h.push( 104 );
printf( "%d\n" , h.top() ); // 104, 104 % 3 == 2
}
</code></pre>
</section>
<section>
<h1>常見問題</h1>
<ul>
<li>Greedy的輔助工具</li>
<li>動態維護極值</li>
</ul>
</section>
<section>
<h2>非負權圖上單源最短路</h2>
<h3>Dijkstra 算法</h3>
</section>
<section>
<pre><code class="cpp" data-trim contenteditable>
vector<PII> e[ N ];
priority_queue<PII,vector<PII>,greater<PII> > heap;
void Dijkstra( int source ){
heap.push( { 0 , source } );
while( heap.size() ){
PII tp = heap.top(); heap.pop();
int tdst = tp.first , tnode = tp.second;
if( vst[ tnode ] ) continue;
dst[ tnode ] = tdst;
vst[ tnode ] = true;
for( size_t i = 0 ; i < e[ tnode ].size() ; i ++ ){
int nxt = e[ tnode ][ i ].first;
int nxtdst = e[ tnode ][ i ].second + dst[ tnode ];
if( vst[ nxt ] ) continue;
heap.push( { nxtdst , nxt } );
}
}
}
</code></pre>
</section>
<section>
<h1>這樣就滿足了嗎？</h1>
</section>
<section>
<h2>Black Magic</h2>
<pre><code class="cpp" data-trim contenteditable>
#include <bits/extc++.h>
#include <stdio.h>
using namespace __gnu_pbds;
typedef priority_queue<int> heap;
int main(){
heap h1 , h2;
h1.push( 1 ); h1.push( 3 );
h2.push( 2 ); h2.push( 4 );
h1.join( h2 );
printf( "%d\n" , (int)h1.size() ); // 4
printf( "%d\n" , (int)h2.size() ); // 0
printf( "%d\n" , h1.top() ); // 4
h1.pop();
}
</code></pre>
<h2>插入及合併$\Theta(1)$!</h2>
</section>
</section>
<!-- disjoint set -->
<section>
<section>
<h1>並查集</h1>
<h1>Disjoint Set</h1>
</section>
<section>
<h2>What is disjoint set?</h2>
<ul>
<li>維護兩兩互斥的集合</li>
<li>支援合併及查詢</li>
</ul>
</section>
<section>
<img src="images/djs.png">
</section>
<section>
<h2>How to do so?</h2>
<h3>沒有現成的可以用QQ</h3>
</section>
<section>
<h2>How to do so?</h2>
<h3>超好寫！</h3>
<pre><code class="cpp" data-trim contenteditable>
struct DisjointSet{
int p[ N ];
void Init( int n ){
for( int i = 1 ; i <= n ; i ++ ) p[ i ] = i;
}
int Find( int x ){
return x == p[ x ] ? x : Find( p[ x ] );
}
void Union( int x , int y ){
p[ Find( x ) ] = Find( y );
}
};
</code></pre>
</section>
<section>
<h1>Upgrade!! 路徑壓縮</h1>
<pre><code class="cpp" data-trim contenteditable>
int Find( int x ){
return x == p[ x ] ? x : p[ x ] = Find( p[ x ] );
}
</code></pre>
<h2>查詢複雜度$O(\lg N)$</h2>
</section>
<section>
<img src="images/djs-1.png">
</section>
<section>
<h1>Upgrade again!! Union by rank</h1>
<h2>啓發式合併的思想</h2>
<h2>查詢複雜度$O(\lg N)$</h2>
</section>
<section>
<pre><code class="cpp" data-trim contenteditable>
int p[ N ] , sz[ N ];
void Init( int n ){
for( int i = 1 ; i <= n ; i ++ )
p[ i ] = i, sz[ i ] = 1;
}
void Union( int x , int y ){
x = Find( x ); y = Find( y );
if( x == y ) return;
p[ x ] = y;
sz[ y ] += sz[ x ];
}
</code></pre>
</section>
<section>
<h1>雙管齊下！</h1>
<h2>操作均攤複雜度$O(\alpha(n))$</h2>
</section>
<section>
<h1>常見問題</h1>
<ul>
<li>連通塊</li>
<li>等價性質</li>
</ul>
</section>
<section>
<p>無向圖 $G$，$N$ 個點、$M$ 條邊，破壞 $Q$ 條邊，問每破壞一條邊後連通塊數？</p>
<img src="images/ural1671.png">
</section>
<section>
<h2>剛剛好像沒有教過將並查集分裂...</h2>
</section>
<section>
<h1>時光倒流！</h1>
<p>破壞 $\rightarrow$ 建造</p>
</section>
<section>
<p>三類動物，$A$ 吃 $B$、$B$ 吃 $C$、$C$ 吃 $A$。給 $N$ 個動物，$K$ 筆資訊包含：</p>
<ul>
<li>$X$ 和 $Y$ 是同類</li>
<li>$X$ 吃 $Y$</li>
</ul>
<p>問有多少條資訊錯誤？</p>
</section>
<section>
<h2>連通塊?</h2>
<h3>$X$ 吃 $Y$ 是同一個集合？</h3>
</section>
<section>
<h2>維護不互斥的資訊！</h2>
<h3>資訊：$X_i$，$X$ 屬於第 $i$ 類</h3>
</section>
<section>
<h2> $X$ 和 $Y$ 同類</h2>
<ol>
<li>檢查是否有 $X_A$ 和 $Y_B,Y_C$ 同類，...</li>
<li>將 $X_A,Y_A$、$X_B,Y_B$、$X_C,Y_C$ 合併爲一個並查集</li>
</ol>
</section>
<section>
<h2> $X$ 吃 $Y$</h2>
<ol>
<li>檢查是否有 $X_A$ 和 $Y_A,Y_C$ 同類，...</li>
<li>將 $X_A,Y_B$、$X_B,Y_C$、$X_C,Y_A$ 合併爲一個並查集</li>
</ol>
</section>
<section>
<pre><code class="cpp" data-trim contenteditable>
if( type == 1 ){
if( djs.Find( trans( x , 0 ) ) == djs.Find( trans( y , 1 ) ) ||
djs.Find( trans( x , 0 ) ) == djs.Find( trans( y , 2 ) ) ){
fake ++; continue;
}
for( int i = 0 ; i < 3 ; i ++ )
djs.Union( trans( x , i ) , trans( y , i ) );
}else{
if( djs.Find( trans( x , 0 ) ) == djs.Find( trans( y , 0 ) ) ||
djs.Find( trans( x , 0 ) ) == djs.Find( trans( y , 2 ) ) ){
fake ++; continue;
}
for( int i = 0 ; i < 3 ; i ++ )
djs.Union( trans( x , i ) , trans( y , ( i + 1 ) % 3 ) );
}
</code></pre>
</section>
<section>
<h1>這樣就滿足了嗎？</h1>
<h3>(again)</h3>
</section>
<section>
<h1>Connected?</h1>
<p>給你 $N$ 個點，$Q$ 筆操作：加邊或將最近 $k$ 條邊拔除，每筆操作完輸出連通塊個數</p>
</section>
<section>
<h1>Connected?</h1>
<ul>
<li>可持久化 Disjoint Set?</li>
<li>倒著做?</li>
</ul>
</section>
<section>
<h1>Undo disjoint set!</h1>
</section>
<section>
<h2>Undo disjoint set!</h2>
<p>Find, Union都是一些assign操作!</p>
</section>
<section>
<h2>Undo disjoint set!</h2>
<pre><code class="cpp" data-trim contenteditable>
vector< pair<int*,int> > h; vector<int> sp;
void assign( int *k, int v ){ h.push_back( {k, *k} ); *k = v; }
void save(){ sp.push_back(h.size()); }
void undo(){
int last=sp.back(); sp.pop_back();
while( h.size()!=last ){
auto x=h.back(); h.pop_back();
*x.first = x.second;
}
}
void Union( int x , int y ){
x = Find( x ); y = Find( y );
if( x == y ) return;
assign( &sz[ x ] , sz[ x ] + sz[ y ] );
assign( &p[ y ] , x );
}
</code></pre>
</section>
</section>
<!-- segment tree -->
<section>
<section>
<h1>線段樹</h1>
<h1>Segment Treee</h1>
</section>
<section>
<p>給你長度 $N$ 的序列 $v$，$Q$ 筆操作：</p>
<ul>
<li>將 $v_i$ 加上 $d$</li>
<li>詢問區間 $[l,r]$ 的總和</li>
</ul>
</section>
<section>
<h2>What is segment tree?</h2>
<ul>
<li>二元樹狀結構</li>
<li>處理區間問題</li>
</ul>
</section>
<section>
<img src="images/st.png">
</section>
<section>
<h2>實作方式-node</h2>
<pre><code class="cpp" data-trim contenteditable>
struct Node{
int val;
Node *lc , *rc;
Node(){ val = 0; lc = rc = NULL; }
void pull(){
val = lc->val + rc->val;
}
};
</code></pre>
</section>
<section>
<h2>初始化線段樹 四部曲</h2>
<ol>
<li>葉節點直接得到資訊</li>
<li>建立左子樹</li>
<li>建立右子樹</li>
<li>pull</li>
</ol>
</section>
<section>
<h2>實作方式-build</h2>
<pre><code class="cpp" data-trim contenteditable>
Node* build( int L , int R ){
Node *node = new Node();
if( L == R ){
node->val = v[ L ];
return node;
}
int mid = ( L + R ) >> 1;
node->lc = build( L , mid );
node->rc = build( mid + 1 , R );
node->pull();
return node;
}
</code></pre>
</section>
<section>
<h2>修改線段樹 三部曲</h2>
<ol>
<li>葉節點直接修改資訊</li>
<li>修改包含 $i$ 的子樹</li>
<li>pull</li>
</ol>
</section>
<section>
<h2>實作方式-modify</h2>
<pre><code class="cpp" data-trim contenteditable>
void modify( Node* node , int L , int R , int i , int d ){
if( L == R ){
assert( L == i );
node->val += d;
return;
}
int mid = ( L + R ) >> 1;
if( i <= mid ) modify( node->lc , L , mid , i , d );
else modify( node->rc , mid + 1 , R , i , d );
node->pull();
}
</code></pre>
</section>
<section>
<img src="images/st-1.png">
</section>
<section>
<h2>詢問線段樹 五步驟</h2>
<ol>
<li>區間不相交，直接回傳</li>
<li>區間包含，回傳節點上資訊</li>
<li>收集左子樹資訊</li>
<li>收集右子樹資訊</li>
<li>結合兩子樹資訊</li>
</ol>
</section>
<section>
<h2>實作方式-query</h2>
<pre><code class="cpp" data-trim contenteditable>
int query( Node* node , int L , int R , int ql , int qr ){
if( ql > R || qr < L ) return 0;
if( ql <= L && R <= qr ) return node->val;
int mid = ( L + R ) >> 1;
return query( node->lc , L , mid , ql , qr ) +
query( node->rc , mid + 1 , R , ql , qr );
}
</code></pre>
</section>
<section>
<img src="images/st-2.png">
</section>
<section>
<p>給你長度 $N$ 的序列 $v$，$Q$ 筆操作：</p>
<ul>
<li>將區間 $[l,r]$ 都加上 $d$</li>
<li>詢問區間 $[l,r]$ 的總和</li>
</ul>
</section>
<section>
<h2>擴充包-線段樹懶標記(Lazy tag)</h2>
<img src="images/fun.jpe">
</section>
<section>
<img src="images/st-3.png">
</section>
<section>
<img src="images/st-4.png">
</section>
<section>
<img src="images/st-5.png">
</section>
<section>
<img src="images/st-6.png">
</section>
<section>
<h2>Push!</h2>
<pre><code class="cpp" data-trim contenteditable>
void push( Node* node , int L , int R ){
if( !node->tag ) return;
if( L != R ){ // check leaf
int mid = ( L + R ) >> 1;
node->lc->tag += node->tag;
node->rc->tag += node->tag;
node->lc->val += node->tag * ( mid - L + 1 );
node->rc->val += node->tag * ( R - mid );
}
node->tag = 0;
}
</code></pre>
</section>
<section>
<h2>線段樹懶標記</h2>
<ul>
<li>修改、詢問時，遇到標記即push!</li>
</ul>
</section>
<section>
<h2>牛刀小試</h2>
<p>$N \times N$($N \le 10^9$) 矩陣初始值爲零，$Q$ 筆操作:</p>
<ul>
<li> 將 $(x,y)$ 座標加上 d </li>
<li> 詢問 $(x_1,y_1)$ 到 $(x_2,y_2) 子矩陣的和。$ </li>
</ul>
</section>
<section>
<h2>離散化？</h2>
<h2>空間爆炸</h2>
</section>
<section>
<h2>擴充包2-copy on write</h2>
</section>
<section>
<h2>用到的點很少，那要用再開！</h2>
<img src="images/stcow.png">
</section>
<section>
<h2>實作方式</h2>
<p>要往下走時先確認有沒有該節點(想想 Trie!)</p>
<p>pull 時確認是否有該節點</p>
<pre><code class="cpp" data-trim contenteditable>
int Val2( Node2* node ){ return node ? node->val : 0; }
void pull2( Node2* node ){
node->val = Val2( node->lc ) + Val2( node->rc );
}
</code></pre>
</section>
<section>
<h2>牛刀小試2</h2>
<p>給長度 $N 的序列 $v$，$Q$ 筆操作:</p>
<ul>
<li> 將 $v_i$ 加上 d </li>
<li> 詢問第 $k$ 筆操作後，區間 $[l,r]$ 的總和 </li>
</ul>
<h3>強迫在線</h3>
</section>
<section>
<h2>擴充包3-可持久化</h2>
</section>
<section>
<h2>一個版本長一棵！</h2>
<img src="images/stper.png">
</section>
<section>
<h2>實作方式</h2>
<p>共同指向共用的節點</p>
<p>修改時新增自己用到的節點</p>
<p>詢問完全一樣！</p>
</section>
<section>
<h2>三刀並用</h2>
<ul>
<li>區間修改 $\rightarrow$ 懶標記</li>
<li>大範圍區間 $\rightarrow$ Copy on write</li>
<li>版本控制 $\rightarrow$ 可持久化</li>
</ul>
</section>
<section>
<h3>區間反轉、區間搬移、區間複製、$\cdots$</h3>
</section>
</section>
<!-- treap -->
<section>
<section>
<h1>樹堆 Treap</h1>
<h3>邁向序列操作大師之路！</h3>
</section>
<section>
<h2>What is Treap?</h2>
<ul>
<li>Tree+Heap</li>
<li>滿足樹性質</li>
<li>滿足堆性質</li>
</ul>
</section>
<section>
<h2>樹性質？</h2>
<ul>
<li>Key $\ge$ 左子節點的 Key</li>
<li>Key $\le$ 右子節點的 Key</li>
</ul>
<img src="images/treap.png">
</section>
<section>
<h2>堆性質？</h2>
<ul>
<li>Pri $\ge$ 左子節點的 Pri</li>
<li>Pri $\ge$ 右子節點的 Pri</li>
</ul>
<img src="images/treap-1.png">
</section>
<section>
<h2>實作方式-Node</h2>
<pre><code class="cpp" data-trim contenteditable>
struct Treap{
Treap *l , *r;
int pri , key , val;
Treap( int _val , int _key ) :
val( _val ) , key( _key ), l( 0 ), r( 0 ), pri( rand() ) {}
};
</code></pre>
</section>
<section>
<h2>實作方式-merge</h2>
<h3>合併三部曲</h3>
<ul>
<li>樹爲空？</li>
<li>誰當根？</li>
<li>要做誰？</li>
</ul>
</section>
<section>
<h2>實作方式-merge</h2>
<pre><code class="cpp" data-trim contenteditable>
Treap* merge( Treap* a , Treap* b ){
if( !a || !b ) return a ? a : b;
if( a->pri > b->pri ){
a->r = merge( a->r , b );
return a;
}else{
b->l = merge( a , b->l );
return b;
}
}
</code></pre>
</section>
<section>
<h2>實作方式-split</h2>
<h3>分裂三部曲</h3>
<ul>
<li>樹爲空？</li>
<li>根送誰？</li>
<li>要做誰？</li>
</ul>
</section>
<section>
<h2>實作方式-split</h2>
<pre><code class="cpp" data-trim contenteditable>
void split( Treap* t , int k , Treap *&a , Treap *&b ){
if( !t ) a = b = NULL;
else if( t->key <= k ){
a = t;
split( t->r , k , a->r , b );
}else{
b = t;
split( t->l , k , a , b->l );
}
}
</code></pre>
</section>
<section>
<h2>沒有其他操作？</h2>
</section>
<section>
<h2>插入操作-Insert</h2>
<pre><code class="cpp" data-trim contenteditable>
Treap* insert( Treap* t , int k ){
Treap *tl , *tr;
split( t , k , tl , tr );
return merge( tl , merge( new Treap( k ) , tr ) );
}
</code></pre>
</section>
<section>
<h2>刪除操作-remove</h2>
<pre><code class="cpp" data-trim contenteditable>
Treap* remove( Treap* t , int k ){
Treap *tl , *tr;
split( t , k - 1 , tl , t );
split( t , k , t , tr );
return merge( tl , tr );
}
</code></pre>
</section>
<section>
<h1>去征服所有序列題吧！</h1>
</section>
</section>
<section>
<h2>Good Luck & Have Fun!</h2>
</section>
</div>
</div>
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