几种求LCA的姿势

 FZSZ Online Judge #815. LCA

题目描述

给定一个 n 个点的有根树,q 次询问 (x,y) 的最近公共祖先

输入格式

一行输入一个正整数 n

以下 n1 行, 每行输入两个正整数 (x,y), 表示 xy有一条边

一行输入一个正整数 q

以下 q 行, 每行输入两个正整数 (x,y), 表示询问 lca(x,y)

输出格式

设第 i 次询问的答案为 xixi, 输出 qi=1nq-ixi(mod264)

//不知道有没搞错反正是 ans = ans * n + Lca(x, y);

样例输入

3
1 2
1 3
3
1 2
1 3
2 3

样例输出

13

数据范围

n,q106,x,yn,

倍增算法(在线处理)

#include<cstdio>
#include<iostream>
#include<cmath>
#define ull unsigned long long
#define BIG 2333333
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
ull mod;
inline int read(){
	char c;
	while(c=getchar(),c>'9'||c<'0');
	int data=c-48;
	while(c=getchar(),c>='0'&&c<='9')
		data=(data<<1)+(data<<3)+c-48;
	return data;
}
int nxt[BIG],las[BIG],to[BIG],rd[BIG],dep[BIG],fa[BIG][24];
int n,q,tot,x,y;
ull ans;
inline void add(int x,int y){
	nxt[++tot]=las[x];
	las[x]=tot;
	to[tot]=y;
	++rd[y];
}
inline void dfs(int now){
	for(register int e=las[now];e;e=nxt[e]){
		dep[to[e]]=dep[now]+1;
		fa[to[e]][0]=now;
		dfs(to[e]);
	}
}
inline void ST(){
	FOR(i,1,20)
		FOR(j,1,n)
			fa[j][i]=fa[fa[j][i-1]][i-1];
}
inline void swap(int &x,int &y){
	int t=x;
	x=y;
	y=t;
}
inline int LCA(int x,int y){
	if(dep[x]<dep[y])
		swap(x,y);
	for(register int i=20;i>=0;--i)
		if(dep[fa[x][i]]>=dep[y])
			x=fa[x][i];
	if(x==y)
		return x;
	for(register int i=20;i>=0;--i)
		if(fa[x][i]!=fa[y][i])
			x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}
int wr[101];
inline void write(unsigned long long a){
	if(!a){
		putchar(48);
		return;
	}
	while(a)
		wr[++wr[0]]=a%10,a/=10;
	while(wr[0])
		putchar(48+wr[wr[0]--]);
	return;
}
int main(){
	n=read();
	FOR(i,1,n-1){
		x=read();
		y=read();
		add(x,y);
	}
	FOR(i,1,n)
		if(!rd[i]){
			dep[i]=1;
			dfs(i);
			break;
		}
	q=read();
	ST();
	mod=1;
	FOR(i,1,63)
		mod=(ull)(mod<<1);
	while(q--){
		x=read();
		y=read();
		ans=ans*n+LCA(x,y);
		if(ans<0){
			ans+=mod;
			ans+=mod;
		}
	}
	write(ans);
	return 0;
}

树链剖分(在线处理)

#include<cstdio>
#include<iostream>
#define BIG 2333333
#define ll unsigned long long 
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
namespace TCP{
	int nxt[BIG],las[BIG],to[BIG],deg[BIG];
	int dep[BIG],sz[BIG],top[BIG],f[BIG];
	int n,m,x,y,tot;
	inline void add(int x,int y){
		nxt[++tot]=las[x];
		las[x]=tot;
		to[tot]=y;
		++deg[y];
		return;
	}
	inline void dfs1(int now){
		sz[now]=1;
		for(register int e=las[now];e;e=nxt[e]){
			dep[to[e]]=dep[f[to[e]]=now]+1;
			dfs1(to[e]);
			sz[now]+=sz[to[e]];
		}
		return;
	}
	inline void dfs2(int now,int chain){
		top[now]=chain;
		register int i=0;
		for(register int e=las[now];e;e=nxt[e])
			if(sz[i]<sz[to[e]])
				i=to[e];
		if(!i)
			return;
		dfs2(i,chain);
		for(register int e=las[now];e;e=nxt[e])
			if(to[e]!=i)
				dfs2(to[e],to[e]);
		return;
	}
	inline ll lca(int u,int v){
		for(;top[u]!=top[v];dep[top[u]]>dep[top[v]]?u=f[top[u]]:v=f[top[v]]);
		return (ll)(dep[u]<dep[v]?u:v);
	}
};
using namespace TCP; 
namespace DATA{
	inline void read(int &x){
		char c;
		while(c=getchar(),c>'9'||c<'0');
		x=c-48;
		while(c=getchar(),c>='0'&&c<='9')
			x=(x<<1)+(x<<3)+c-48;
		return;
	}
	int wr[51]; 
	inline void write(ll x){
		while(x)
			wr[++wr[0]]=x%10,x/=10;
		while(wr[0])
			putchar(48+wr[wr[0]--]);
		putchar('
');
		return;
	}
};
using namespace DATA;
namespace WORK{
	ll ans,mod=1ll;
	inline void work(){
		read(n);
		FOR(i,1,n-1){
			read(x);
			read(y);
			add(x,y);
		}
		FOR(i,1,n)
			if(!deg[i]){
				dfs1(i);
				dfs2(i,i);
				break;
			}
		FOR(i,1,63)
			mod<<=1;
		read(m);
		while(m--){
			read(x);
			read(y);
			ans=ans*n+lca(x,y);
			if(ans<0){
				ans+=mod;
				ans+=mod;
			}
		}
		write(ans); 
	} 
}; 
using namespace WORK;
int main(){
	work();
	return 0;
} 

Tarjan(离线处理)

#include<cstdio>
#include<iostream>
#include<vector>
#define BIG 2333333
#define ll unsigned long long 
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
namespace TARJAN{
	int n,m,tot;
	int x,y,z;
	int las[BIG],to[BIG],nxt[BIG],f[BIG],used[BIG],deg[BIG];
	class question{
		public:
		int u[3];
		int lca;
		private:
	}q[BIG];
	vector<int>point[BIG>>1];
	inline void add(int x,int y){
		nxt[++tot]=las[x];
		las[x]=tot;
		to[tot]=y;
		++deg[y];
		return;
	}
	inline int find(int x){
		return f[x]==x?x:f[x]=find(f[x]);
	}
	inline void dfs(int now,int fa){
		used[now]=1;
		register int i,p,other;
		for(i=0;i<(int)point[now].size();++i){
			p=point[now][i];
			for(register int j=0;j<=1;++j){	
				other=q[p].u[j];
				if(other!=now)
					if(used[other])
						q[p].lca=find(other);
			}
		}
		for(register int e=las[now];e;e=nxt[e])
			dfs(to[e],now);
		f[now]=fa;
		return;
	}
}
using namespace TARJAN; 
namespace DATA{
	inline void read(int &x){
		char c;
		while(c=getchar(),c>'9'||c<'0');
		x=c-48;
		while(c=getchar(),c>='0'&&c<='9')
			x=(x<<1)+(x<<3)+c-48;
		return;
	}
	int wr[51]; 
	inline void write(ll x){
		while(x)
			wr[++wr[0]]=x%10,x/=10;
		while(wr[0])
			putchar(48+wr[wr[0]--]);
		putchar('
');
		return;
	}
};
using namespace DATA;
namespace WORK{
	ll ans,mod=1ll;
	inline void work(){
		read(n);
		FOR(i,1,n-1){
			read(x);
			read(y);
			add(x,y);
		}
		FOR(i,1,n)
			f[i]=i;
		FOR(i,1,63)
			mod<<=1;
		read(m);
		FOR(i,1,m){
			read(q[i].u[0]);
			read(q[i].u[1]);
			point[q[i].u[0]].push_back(i);
			point[q[i].u[1]].push_back(i); 
		}
		FOR(i,1,n)
			if(!deg[i]){
				dfs(i,0);
				break;
			}
		FOR(i,1,m){
			ans=(ll)(ans*n+q[i].lca);
			if(ans<0){
				ans+=mod;
				ans+=mod;
			}
		}
		write(ans);
		return;
	} 
};
using namespace WORK;
int main(){
	work();
	return 0;
} 

  

 

原文地址:https://www.cnblogs.com/Stump/p/7687048.html