BZOJ3083: 遥远的国度

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BZOJ100题辣（已经无法直视的正确率

树剖板子题，注意和dfs序结合，根据根的变化变换统计的方式即可。

//BZOJ 3083
//by Cydiater
//2016.10.23
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
#include <ctime>
#include <iomanip>
#include <cstdlib>
#include <cstdio>
#include <cmath>
using namespace std;
#define ll long long
#define up(i,j,n)		for(  int i=j;i<=n;i++)
#define down(i,j,n)		for(int i=j;i>=n;i--)
#define FILE "bbbbb"
const int MAXN=1e5+5;
const ll oo=1LL<<32;
inline ll read(){
	char ch=getchar();ll x=0,f=1;
	while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
	return x*f;
}
int N,M,dep[MAXN],LINK[MAXN],len=0,fa[MAXN][25],son[MAXN],siz[MAXN],top[MAXN],seg[MAXN],cnt=0,pos[MAXN],ROOT,opt,L,R;
ll v,pro[MAXN];
int c=0;
struct edge{
	int y,next;
}e[MAXN<<1];
struct Tree{
	ll v,delta;
}t[MAXN<<3];
namespace solution{
	inline void insert(int x,int y){e[++len].next=LINK[x];LINK[x]=len;e[len].y=y;}
	inline void reload(int root){t[root].v=min(t[root<<1].v,t[root<<1|1].v);}
	inline void downit(int root){
		if(t[root].delta==0)return;
		int delta=t[root].delta;t[root].delta=0;
		t[root<<1].delta=delta;t[root<<1].v=delta;
		t[root<<1|1].delta=delta;t[root<<1|1].v=delta;
	}
	void init(){
		N=read();M=read();
		up(i,2,N){
			int x=read(),y=read();
			insert(x,y);
			insert(y,x);
		}
		up(i,1,N)pro[i]=read();
		ROOT=read();
	}
	void dfs1(int node,int deep,int father){
		fa[node][0]=father;dep[node]=deep;son[node]=0;
		siz[node]=1;int max_siz=0;
		for(int i=LINK[node];i;i=e[i].next)if(e[i].y!=father){
			dfs1(e[i].y,deep+1,node);
			siz[node]+=siz[e[i].y];
			if(siz[e[i].y]>max_siz){
				max_siz=siz[e[i].y];
				son[node]=e[i].y;
			}
		}
	}
	void dfs2(int node,int TOP){
		top[node]=TOP;seg[++cnt]=node;pos[node]=cnt;
		if(son[node])dfs2(son[node],TOP);
		for(int i=LINK[node];i;i=e[i].next)if(e[i].y!=fa[node][0]&&e[i].y!=son[node])
			dfs2(e[i].y,e[i].y);
	}
	void get_ancestor(){
		up(i,1,21)up(node,1,N)if(fa[node][i-1]!=0)
			fa[node][i]=fa[fa[node][i-1]][i-1];
	}
	void build(int leftt,int rightt,int root){
		if(leftt==rightt){
			t[root].delta=0;
			t[root].v=pro[seg[leftt]];
			return;
		}
		int mid=(leftt+rightt)>>1;
		build(leftt,mid,root<<1);
		build(mid+1,rightt,root<<1|1);
		reload(root);
	}
	void Build(){
		dfs1(ROOT,0,0);dfs2(ROOT,ROOT);
		get_ancestor();
		build(1,N,1);
	}
	int LCA(int x,int y){
		if(x==y)	return x;
		if(dep[x]<dep[y])swap(x,y);
		down(i,21,0)if(dep[x]-(1<<i)>=dep[y])x=fa[x][i];
		if(x==y)	return x;
		down(i,21,0)if(fa[x][i]!=0&&fa[x][i]!=fa[y][i]){
			x=fa[x][i];
			y=fa[y][i];
		}
		return fa[x][0];
	}
	void updata(int leftt,int rightt,int root){
		downit(root);
		if(leftt>R||rightt<L)	return;
		if(leftt>=L&&rightt<=R){
			t[root].delta=t[root].v=v;
			return;
		}
		int mid=(leftt+rightt)>>1;
		updata(leftt,mid,root<<1);
		updata(mid+1,rightt,root<<1|1);
		reload(root);
	}
	void change(int x,int aim){
		while(top[x]!=top[aim]){
			R=pos[x];L=pos[top[x]];
			updata(1,N,1);
			x=fa[top[x]][0];
		}
		R=pos[x];L=pos[aim];
		updata(1,N,1);
	}
	void Change(int x,int y){
		int lca=LCA(x,y);
		change(x,lca);change(y,lca);
	}
	ll get(int leftt,int rightt,int root){
		downit(root);
		if(leftt>R||rightt<L)		return oo;
		if(leftt>=L&&rightt<=R)		return t[root].v;
		int mid=(leftt+rightt)>>1;
		return min(get(leftt,mid,root<<1),get(mid+1,rightt,root<<1|1));
	}
	void slove(){
		while(M--){
			opt=read();
			if(opt==1)ROOT=read();
			else if(opt==2){
				int x=read(),y=read();v=read();
				Change(x,y);
			}else{
				int node=read(),Pos=pos[ROOT];ll ans;
				int leftt=pos[node],rightt=leftt+siz[node]-1;
				if(Pos==leftt){
					L=1;R=N;
					ans=get(1,N,1);
				}else if(!(Pos>=leftt&&Pos<=rightt)){
					L=leftt;R=rightt;
					ans=get(1,N,1);
				}else{
					for(int i=LINK[node];i;i=e[i].next)if(e[i].y!=fa[node][0]){
						leftt=pos[e[i].y];rightt=pos[e[i].y]+siz[e[i].y]-1;
						if(Pos>=leftt&&Pos<=rightt){
							L=1;R=leftt-1;
							if(R>=L)ans=get(1,N,1);
							L=rightt+1,R=N;
							if(L<=R)ans=min(ans,get(1,N,1));
							break;
						}
					}
				}
				printf("%lld\n",ans);
			}
		}
	}
}
int main(){
	//freopen(FILE".in","r",stdin);
	//freopen(FILE".out","w",stdout);
	//freopen("input.in","r",stdin);
	//freopen("out.out","w",stdout);
	using namespace solution;
	init();
	Build();
	slove();
	return 0;
}

BZOJ3083: 遥远的国度

标签：amp   cst   print   load   lld   targe   online   long   stdin   

原文地址：http://www.cnblogs.com/Cydiater/p/5989916.html