L2-012 Judgment about heap (simulated heap + string processing)

topic link

https://pintia.cn/problem-sets/994805046380707840/problems/994805064676261888

ideas

We first implement the heap by simulating (here we only need to implement the insertion of the heap), and then for four cases:

• x is the root node, we only need to determine whether the first element in the heap is x
• x is the parent node of y, we only need to judge $Y>>Is\; 1$ equal to$x$ because the root node starts from 1, the same is true below
• x is a child node of y, we only need to judge $x>>Is\; 1$ equal to$Y$
• x and y are siblings, we just need to judge $(x>>1)==(and>>1)$

Note that it is very possible to think too much here, such as whether the child node is considered a grandchild node (not counted here), whether the sibling node is a half-brother (also not counted here), and then it is wrong. For more details, please see code

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000009
#define endl "\n"
#define PII pair<int,int>

const int N = 2e6+10;
int n,m,cnt,h[N],hp[N],ph[N];
int k,x,kk=0;
//ph记录的是第i次存入的下标
//hp记录的是下标为i是第几次存入的
//ph和hp是一个互逆的数组

void heap_swap(int a,int b){
    
    //a、b都是下标
	swap(ph[hp[a]],ph[hp[b]]);
	swap(hp[a],hp[b]);
	swap(h[a],h[b]);
}
void up(int u) {
    
    //向上更新
	while((u>>1) && h[u] < h[u>>1]) {
    
    
		heap_swap(u,u >> 1);
		u >>= 1;
	}
}
void insert(int k){
    
    
	cnt++;
	kk++;//表示是第kk次插入的
	h[cnt] = k;
	ph[kk] = cnt,hp[cnt] = kk;
	up(cnt);
}

map<int,int> st;

int main()
{
    
    
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	string op;
	cin>>n>>m;
	
	for(int i = 1;i <= n; ++i){
    
    
		cin>>k;
		insert(k);
	} 
	for(int i = 1;i <= n; ++i)
		st[h[i]] = i;
	
	while(m--){
    
    
		string op;
		int x,y;
		cin>>x>>op;
		bool ans;
		if(op == "is"){
    
    
			cin>>op;
			if(op == "the") {
    
    
				cin>>op;
				if(op == "root") {
    
    //x是根结点
					if(h[1] == x) ans = true;
					else ans = false;
				} else {
    
    //x是y的父结点
					cin>>op>>y;
					x = st[x],y = st[y];
					if(x < y && (y>>1 == x)) ans = true;
					else ans = false;
				}
			} else {
    
    //x是y的一个子结点
				cin>>op>>op>>y;
				x = st[x],y = st[y];
				if(x > y && (x>>1 == y)) ans = true;
				else ans = false;
			}
		} else {
    
    //x和y是兄弟结点
			cin>>y;
			getline(cin,op);
			x = st[x],y = st[y];
			if((x>>1) == (y>>1)) ans = true;
			else ans = false;
		}
		if(ans) cout<<"T"<<endl;
		else cout<<"F"<<endl;
	}
	
	
	return 0;
}