[BZOJ4538]网络

今天打比赛，毒瘤yww把这题出到$n,m\leq 5\times10^5$，因为不会写整体二分所以来写坑爹的$O\left(n\log_2n\right)$做法

考虑按重要度建权值线段树（相同权值的请求视作不同的），每个线段树节点存这个区间内的链的交集

那么插入删除就直接搞，询问时在线段树上贪心，如果当前节点的右儿子有请求且它的链交集不覆盖询问点，那么往右儿子走，否则往左儿子走

链的交集还是链，我们要求$(a,b)$和$(c,d)$的交集，如果$lca_{a,b}$在$(c,d)$上，那么一定有交集，交集为$(\text{low}({lca_{c,a}},lca_{c,b}),\text{low}(lca_{d,a},lca_{d,b}))$（其中$\text{low}(a,b)$表示$a,b$中的较低点），反过来也是一样的

细节有点多，为了跑得更快，求lca的部分可以转成rmq然后用ST表$O(1)$查询

#include<stdio.h>#include<map>using namespace std;int h[100010],nex[200010],to[200010],fa[100010],dep[100010],fir[100010],dfn[200010],mp[200010][20],log2[200010],M;void add(int a,int b){M++;to[M]=b;nex[M]=h[a];h[a]=M;}void dfs(int x){M++;dfn[M]=x;fir[x]=M;for(int i=h[x];i;i=nex[i]){if(to[i]!=fa[x]){dep[to[i]]=dep[x]+1;fa[to[i]]=x;dfs(to[i]);M++;dfn[M]=x;}}}int querymin(int l,int r){int k=log2[r-l+1];if(dep[dfn[mp[l][k]]]<dep[dfn[mp[r-(1<<k)+1][k]]])return mp[l][k];elsereturn mp[r-(1<<k)+1][k];}int lca(int x,int y){if(fir[x]>fir[y])swap(x,y);return dfn[querymin(fir[x],fir[y])];}struct ask{int op,x,y,v,id;}q[200010];bool operator<(ask a,ask b){if(a.v==b.v)return a.id<b.id;return a.v<b.v;}map<ask,int>pos;map<ask,int>::iterator it;int s[800010],lp[800010],rp[800010],rv[200010];int dis(int x,int y){return dep[x]+dep[y]-2*dep[lca(x,y)];}bool inc(int x,int y,int u){return dis(x,u)+dis(u,y)==dis(x,y);}void pushup(int x){if(s[x<<1]==0||s[x<<1|1]==0){lp[x]=lp[x<<1]|lp[x<<1|1];rp[x]=rp[x<<1]|rp[x<<1|1];return;}if(lp[x<<1]==0||lp[x<<1|1]==0){lp[x]=rp[x]=0;return;}int a,b,c,d,e,f,h,i;a=lp[x<<1];b=rp[x<<1];c=lca(a,b);d=lp[x<<1|1];e=rp[x<<1|1];f=lca(d,e);if(inc(a,b,f)){h=lca(d,a);i=lca(d,b);lp[x]=dep[h]>dep[i]?h:i;h=lca(e,a);i=lca(e,b);rp[x]=dep[h]>dep[i]?h:i;return;}if(inc(d,e,c)){h=lca(a,d);i=lca(a,e);lp[x]=dep[h]>dep[i]?h:i;h=lca(b,d);i=lca(b,e);rp[x]=dep[h]>dep[i]?h:i;return;}lp[x]=rp[x]=0;}void insert(int p,int L,int R,int l,int r,int x){s[x]++;if(l==r){lp[x]=L;rp[x]=R;return;}int mid=(l+r)>>1;if(p<=mid)insert(p,L,R,l,mid,x<<1);elseinsert(p,L,R,mid+1,r,x<<1|1);pushup(x);}void erase(int p,int l,int r,int x){s[x]--;if(l==r){lp[x]=rp[x]=0;return;}int mid=(l+r)>>1;if(p<=mid)erase(p,l,mid,x<<1);elseerase(p,mid+1,r,x<<1|1);pushup(x);}int query(int u,int l,int r,int x){if(s[x]==0)return-1;if(l==r)return inc(lp[x],rp[x],u)?-1:rv[l];int mid=(l+r)>>1;if(s[x<<1|1]!=0&&(lp[x<<1|1]==0||!inc(lp[x<<1|1],rp[x<<1|1],u)))return query(u,mid+1,r,x<<1|1);elsereturn query(u,l,mid,x<<1);}int main(){int n,m,i,j,x,y;scanf("%d%d",&n,&m);for(i=1;i<n;i++){scanf("%d%d",&x,&y);add(x,y);add(y,x);}M=0;dfs(1);for(i=2;i<=M;i++)log2[i]=log2[i>>1]+1;for(i=1;i<=M;i++)mp[i][0]=i;for(j=1;j<20;j++){for(i=1;i<=M;i++){if(i+(1<<j)-1<=M){if(dep[dfn[mp[i][j-1]]]<dep[dfn[mp[i+(1<<(j-1))][j-1]]])mp[i][j]=mp[i][j-1];elsemp[i][j]=mp[i+(1<<(j-1))][j-1];}}}for(i=1;i<=m;i++){scanf("%d%d",&q[i].op,&q[i].x);if(q[i].op==0){scanf("%d%d",&q[i].y,&q[i].v);q[i].id=i;pos[q[i]]=1;}}M=1;for(it=pos.begin();it!=pos.end();it++,M++){it->second=M;rv[M]=(it->first).v;}M--;for(i=1;i<=m;i++){if(q[i].op==0)insert(pos[q[i]],q[i].x,q[i].y,1,M,1);if(q[i].op==1)erase(pos[q[q[i].x]],1,M,1);if(q[i].op==2)printf("%d\n",query(q[i].x,1,M,1));}}