hihocoder #1069 线段树-优快云博客

本文介绍了将树结构转化为数组，随后应用裸线段树进行优化，通过实例展示了数据处理效率的显著提升。利用C++实现，通过代码解析和具体例子阐述了线段树在特定场景下的高效应用。

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学习了树转数组，然后就是一个裸的线段树，或许是数据问题，在线算法比#1067 快了不少

#include<iostream>  
#include<map>
#include<string.h>   
#include<algorithm>  
#include<fstream>
#include<cmath>  
#include<vector>
#include<queue>
#include<map>
using namespace std;  
#define lch(i) ((i)<<1)  
#define rch(i) ((i)<<1|1)  
#define sqr(i) ((i)*(i))  
#define pii pair<int,int>  
#define mp make_pair  
#define FOR(i,b,e) for(int i=b;i<=e;i++)  
#define FORE(i,b,e) for(int i=b;i>=e;i--)  
#define ms(a)   memset(a,0,sizeof(a))  
const int maxnum =200005;
const int mod = 1000000007;
int n,m,len,deepest;


struct node
{
	char*					name;
	int						pans;
	int						depth;
	vector<node*>			son;
	node():					name(NULL),
							depth(0),
							son(),
							pans(0){}
	node(char *str):		depth(0),
							son(),
							pans(0){
		name = new char[strlen(str)];
		strcpy(name,str);
	}
	~node(){
		if(name)
			delete[] name;
	}
};
node* root;
map<string,node*> name2node;
node* ans[maxnum*2];int num=0;
struct tnode
{
	int l,r;
	node* minn;
	int gmin(){
		return (l+r)>>1;
	}
}tree[maxnum*2];

void dfs(node* rt,int dep){
	rt->depth=dep;
	ans[++num]=rt;
	int si = rt->son.size();
	FOR(i,0,si-1){
		dfs(rt->son[i],dep+1);
		ans[++num]=rt;
	}
	rt->pans = num;
}

void Btree(int rt,int l,int r){
	tree[rt].l=l;
	tree[rt].r=r;
	tree[rt].minn = ans[l];
	if(l==r)return;
	int mid = tree[rt].gmin();
	Btree(lch(rt),l,mid);
	Btree(rch(rt),mid+1,r);
	tree[rt].minn=tree[lch(rt)].minn;
	if(tree[rch(rt)].minn->depth<tree[lch(rt)].minn->depth)
		tree[rt].minn=tree[rch(rt)].minn;
	return;
}

node* Stree(int rt,int l,int r){
	if(l==tree[rt].l&&tree[rt].r==r){
		return tree[rt].minn;
	}
	int mid = tree[rt].gmin();
	if(l>mid)
		return Stree(rch(rt),l,r);
	else if(r<=mid)
		return Stree(lch(rt),l,r);
	node* tmp1 =Stree(lch(rt),l,mid);
	node* tmp2 = Stree(rch(rt),mid+1,r);
	return tmp1->depth<tmp2->depth?tmp1:tmp2;
		
}

int main()    
{  
	/*fstream fin("G:/1.txt");
	fin>>n;*/
	cin>>n;
	char name[1000];
	root=0;
	node *far,*son;
	FOR(i,1,n){
		scanf("%s",name);
		/*fin>>name;*/
		if(name2node[name]==0){
			far = new node(name);
			name2node[name]=far;
		}
		else far = name2node[name];
		if(root==0)
			root = far;
		scanf("%s",name);
		/*fin>>name;*/
		if(name2node[name]==0){
			son = new node(name);
			name2node[name]=son;
		}else
			far = name2node[name];
		far->son.push_back(son);
	}

	dfs(root,0);
	Btree(1,1,num);
	cin>>m;
	/*fin>>m;*/
	int st,en;
	FOR(i,1,m){
		scanf("%s",name);
	/*	fin>>name;*/
		far = name2node[name];
		scanf("%s",name);
		/*fin>>name;*/
		son = name2node[name];
		st = far->pans;
		en = son->pans;
		if(st>en) swap(st,en);
		printf("%s\n",Stree(1,st,en)->name);
	}
	
	
	return 0;
}