AVL树模板

 

今天刷PAT甲级的时候遇到了AVL的模板题,之前一直都不会,然后看到别人的博客,终于看懂了.

2021.3.9更:

老的那个太丑了,自己写了个简短好看的

题目链接:https://pintia.cn/problem-sets/994805342720868352/problems/994805404939173888

博客链接:https://www.cnblogs.com/chenxiwenruo/p/6806292.html

看懂之后自己修改成了自己熟悉的数据存储方法(数组)实现了一下,

代码:

#include<iostream>
#include<string.h>
#include<cmath>
#include<set>
#include<map>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<algorithm>
#include<climits>
using namespace std;
typedef long long ll;
inline int read() {
    int sum = 0, f = 1;
    char p = getchar();
    for (; !isdigit(p); p = getchar()) if (p == '-')f = -1;
    for (; isdigit(p); p = getchar())  sum = sum * 10 + p - 48;
    return sum * f;
}
const int maxn = 10001;
#define ls(x) tree[x][0]
#define rs(x) tree[x][1]
int n;
int tree[maxn][2], tot,val[maxn],hight[maxn];
void push_up(int root) {
    hight[root] = max(hight[ls(root)], hight[rs(root)]) + 1;
}
void newnode(int& x,int w) {
    x = ++tot;
    hight[x] = 1;
    val[x] = w;
}
void LL(int &root) {
    int x = root, y = ls(x);
    ls(x) = rs(y);
    rs(y) = x;
    push_up(x);
    push_up(y);
    root = y;
}
void RR(int& root) {
    int x = root, y = rs(x);
    rs(x) = ls(y);
    ls(y) = x;
    push_up(x);
    push_up(y);
    root = y;
}
void LR(int&root) {
    RR(ls(root));
    LL(root);
}
void RL(int& root) {
    LL(rs(root));
    RR(root);
}
void insert(int &root,int w){
    if (root == 0) {
        newnode(root, w); return;
    }
    if (val[root] <= w) {
        insert(rs(root), w);
        if (hight[rs(root)] - hight[ls(root)] > 1) {
            if (val[rs(root)] > w) RL(root);
            else RR(root);
        }
    }
    else {
        insert(ls(root), w);
        if (hight[ls(root)] - hight[rs(root)] > 1) {
            if (val[ls(root)] < w)LR(root);
            else LL(root);
        }
    }
    push_up(root);
}
int main() {
    n = read();
    int root = 0,t;
    for (int i = 1; i <= n; i++) {
        t = read();
        insert(root, t);
    }
    cout << val[root] << endl;
    return 0;
}