poj3694 Network Tarjan+树链剖分

嗯就如上一篇所说，我们缩完点之后，就成为了一棵树，然后每个点权对应它到他的父节点的边是否狗带，然后每次LCA在链上乱跳的时候维护下清空标记就行了，qlog^2n的果然跑得快，172ms。。。。。。

Problem: 3694		User: BPM136
Memory: 19208K		Time: 172MS
Language: G++		Result: Accepted
Source Code
/* ***********************************************
Author        :BPM136
Created Time  :2016/5/6 22:20:26
File Name     :A.cpp
************************************************ */

#include
#include
#include
#include
#include
#include
#define LL long long
#define get(a,i) a&(1<<(i-1))
#define PAU putchar(32)
#define ENT putchar(10)
#define fo(_i,_a,_b) for(int _i=_a;_i<=_b;_i++)
#define efo(_i,_a) for(int _i=last[_a];_i!=0;_i=e[_i].next)
#define file(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
#define setlargestack(x) int size=x<<20;char *p=(char*)malloc(size)+size;__asm__("movl %0, %%esp\n" :: "r"(p));
using namespace std;
LL read()
{
         LL f=1,d=0;char s=getchar();
         while (s<48||s>57){if (s==45) f=-1;s=getchar();}
         while (s>=48&&s<=57){d=d*10+s-48;s=getchar();}
         return f*d;
}
LL readln()
{
       LL f=1,d=0;char s=getchar();
       while (s<48||s>57){if (s==45) f=-1;s=getchar();}
       while (s>=48&&s<=57){d=d*10+s-48;s=getchar();}
       while (s!=10) s=getchar();
       return f*d;
}
inline void write(LL x)
{
    if(x==0){putchar(48);return;}if(x<0)putchar(45),x=-x;
    int len=0,buf[20];while(x)buf[len++]=x%10,x/=10;
    for(int i=len-1;i>=0;i--)putchar(buf[i]+48);return;
}
inline void writeln(LL x){write(x);ENT;}

const int N = 100005;
const int M = 200005;

struct edge {
	int y,next;
}ee[M*2],e[M*2];
int last1[N],ne1;
int last[N],ne;

#define efo1(i,x) for(int i=last1[x];i!=0;i=ee[i].next)

int low[N],dfn[N],Dfn[N],fa[N],mark[N],f[N];

int sum[N<<2],lz[N<<2];

int siz[N],son[N],top[N],first[N],en[N],fat[N],ft[N],dep[N],nu;

int B[M][2],Lb[N],Le[N],L[N],dl[N],Ln,bn;

int n,m,ans,index;

int find(int x) {
	if(f[x]==x) return x;
	return f[x]=find(f[x]);
}

int findt(int x) {
	if(ft[x]==x) return x;
	return ft[x]=findt(ft[x]);
}

void add(int x,int y) {
	ee[++ne1].y=y;ee[ne1].next=last1[x];last1[x]=ne1;
}
void add2(int x,int y) {
	add(x,y); add(y,x);
}

void addt(int x,int y) {
	e[++ne].y=y;e[ne].next=last[x];last[x]=ne;
}

void init() {
	memset(mark,0,sizeof(mark));
	memset(low,0,sizeof(low));
	memset(last1,0,sizeof(last1));
	memset(last,0,sizeof(last));
	memset(fat,0,sizeof(fat));
	memset(dfn,0,sizeof(dfn));
	ne=ne1=ans=index=nu=0; Ln=1; bn=0;
	fo(i,1,n)f[i]=i;
	fo(i,1,m) {
		int x=read(),y=read();
		add2(x,y);
	}
}

void Tarjan(int x) {
	Dfn[x]=Dfn[fa[x]]+1;
	dfn[x]=low[x]=++index;
	efo1(i,x) {
		int y=ee[i].y;
		if(!dfn[y]) {
			fa[y]=x;
			Tarjan(y);
			low[x]=min(low[x],low[y]);
			if(low[y]>dfn[x]) {
				mark[y]=1;
				ans++;
				B[++bn][0]=x,B[bn][1]=y;
			}else f[find(x)]=find(y);
		}else if(fa[x]!=y) low[x]=min(low[x],dfn[y]);
	}
}

int root;

void dfs1(int x,int de) {
	dep[x]=de;siz[x]=1;int mx=0,num=-1;
	efo(i,x) {
		int y=e[i].y;
		if(y==fat[x]) continue;
		fat[y]=x;
		dfs1(y,de+1);
		siz[x]+=siz[y];
		if(siz[y]>mx) {
			mx=siz[y];
			num=y;
		}
	}
	son[x]=num;
}

void dfs2(int x,int t) {
	first[x]=++nu,top[x]=t;
	if(son[x]>0)dfs2(son[x],t); else Le[dl[t]]=nu;
	efo(i,x) {
		int y=e[i].y;
		if(y==fat[x]||y==son[x])continue;
		fat[y]=x;
		dl[y]=++Ln; Lb[dl[y]]=nu+1; 
		dfs2(y,y); //w[y]=1;
	}
	en[x]=nu;
}

#define lch (k<<1)
#define rch (k<<1|1)

void buildsegtree(int k,int l,int r) {
	if(l==r) {
		if(l!=1)sum[k]=1;else sum[k]=0;
		lz[k]=0;
//		cerr<>1;
	buildsegtree(lch,l,mid); buildsegtree(rch,mid+1,r);
	sum[k]=sum[lch]+sum[rch]; lz[k]=0;
//	cerr<>1;
	if(rr<=mid)insert(lch,l,mid,ll,rr);else
	if(ll>mid)insert(rch,mid+1,r,ll,rr);else {
		insert(lch,l,mid,ll,mid);
		insert(rch,mid+1,r,mid+1,rr);
	}
	pushup(k);
}
*/

void pushdown(int k) {
/*
	if(lz[k]==1) {
		int mid=(l+r)>>1;
		lz[lch]=lz[rch]=1;
		sum[lch]=mid-l; sum[rch]=r-mid-1;
	}else if(lz[k]==2){
		sum[lch]=sum[rch]=0;
		lz[lch]=lz[rch]=2;
	}
*/
	sum[lch]=sum[rch]=0;
	lz[lch]=lz[rch]=1;
	lz[k]=0;
}

int ret1;
int get_ans(int k,int l,int r,int _l,int _r) {
	if(l>r||_l>_r)return 0;
	if(l==_l&&r==_r) {
		ret1=sum[k];
		sum[k]=0; lz[k]=1;
		return ret1;
	}
	if(lz[k])pushdown(k);
	int mid=(l+r)>>1;
	int ret=0;
	if(_r<=mid) ret=get_ans(lch,l,mid,_l,_r); else
	if(_l>mid)  ret=get_ans(rch,mid+1,r,_l,_r); else {
		ret=get_ans(lch,l,mid,_l,mid)+get_ans(rch,mid+1,r,mid+1,_r);
	}
//	cout<