BZOJ 4129 树上带修莫队+线段树

思路：
可以先做做BZOJ3585 是序列上的mex
考虑莫队的转移 如果当前数字出现过 线段树上把它置成1
对于询问 二分ans 线段树上查 0到ans的和 是不是ans+1
本题就是把它搞到了序列上 带了个修改…
麻烦一点 本质上是一样的

//By SiriusRen#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N=55555;int n,m,cnt=1,Block,block[N],op,xx,yy,a[N],vis[N],last[N],sum[N],tree[N*8];int first[N],next[N*2],v[N*2],tot,s[N],top,fa[N][20],cnt1,cnt2,deep[N],Ans[N];void add(int x,int y){v[tot]=y,next[tot]=first[x],first[x]=tot++;}void dfs(int x){    for(int i=first[x];~i;i=next[i])        if(v[i]!=fa[x][0])            deep[v[i]]=deep[x]+1,fa[v[i]][0]=x,dfs(v[i]);    s[++top]=x;    if(top==Block){        for(int i=1;i<=top;i++)block[s[i]]=cnt;        cnt++,top=0;    }}struct Ask{    int time,id,l,r;Ask(){}    Ask(int T,int I,int L,int R){time=T,id=I,l=L,r=R;}    friend bool operator<(Ask a,Ask b){        if(block[a.l]==block[b.l]){            if(block[a.r]==block[b.r])return a.time<b.time;            return block[a.r]<block[b.r];        }return block[a.l]<block[b.l];    }}ask[N];struct Change{    int last,position,num;Change(){}    Change(int L,int P,int nn){last=L,position=P,num=nn;}}change[N];void insert(int l,int r,int pos,int num,int f){    if(l==r){tree[pos]=f;return;}    int mid=(l+r)>>1,lson=pos<<1,rson=pos<<1|1;    if(mid<num)insert(mid+1,r,rson,num,f);    else insert(l,mid,lson,num,f);    tree[pos]=tree[lson]+tree[rson];}int query(int l,int r,int pos,int L,int R){    if(l>=L&&r<=R)return tree[pos];    int mid=(l+r)>>1,lson=pos<<1,rson=pos<<1|1;    if(mid<L)return query(mid+1,r,rson,L,R);    else if(mid>=R)return query(l,mid,lson,L,R);    else return query(l,mid,lson,L,R)+query(mid+1,r,rson,L,R);}void reverse(int x){    vis[x]^=1;    if(a[x]<=n){        if(vis[x]){            sum[a[x]]++;            if(sum[a[x]]==1)insert(0,n,1,a[x],1);        }        else{            sum[a[x]]--;            if(!sum[a[x]])insert(0,n,1,a[x],0);        }    }}void change_color(int x,int y){    if(vis[x])reverse(x),a[x]=y,reverse(x);    else a[x]=y;}void work(int x,int y){    while(x!=y){        if(deep[x]<deep[y])swap(x,y);        reverse(x),x=fa[x][0];    }}int lca(int x,int y){    if(deep[x]<deep[y])swap(x,y);    for(int i=19;i>=0;i--)if(deep[x]-(1<<i)>=deep[y])x=fa[x][i];    if(x==y)return x;    for(int i=19;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];    return fa[x][0];}bool check(int x){    return query(0,n,1,0,x)==x+1;}int main(){    memset(first,-1,sizeof(first));    scanf("%d%d",&n,&m),Block=sqrt(n);    for(int i=1;i<=n;i++)scanf("%d",&a[i]),last[i]=a[i];    for(int i=1;i<n;i++)scanf("%d%d",&xx,&yy),add(xx,yy),add(yy,xx);    dfs(1);    for(int i=1;i<=top;i++)block[s[i]]=cnt;    for(int j=1;j<=19;j++)        for(int i=1;i<=n;i++)            fa[i][j]=fa[fa[i][j-1]][j-1];    while(m--){        scanf("%d%d%d",&op,&xx,&yy);        if(op==1){            if(block[xx]>block[yy])swap(xx,yy);            ask[++cnt1]=Ask(cnt2,cnt1,xx,yy);        }        else change[++cnt2]=Change(last[xx],xx,yy),last[xx]=yy;    }    sort(ask+1,ask+1+cnt1);    for(int i=1,T=0;i<=cnt1;i++){        for(;T<ask[i].time;T++){            change_color(change[T+1].position,change[T+1].num);            a[change[T+1].position]=change[T+1].num;        }        for(;T>ask[i].time;T--){            change_color(change[T].position,change[T].last);            a[change[T].position]=change[T].last;        }        if(i!=1)work(ask[i-1].l,ask[i].l),work(ask[i-1].r,ask[i].r);        else work(ask[i].l,ask[i].r);        reverse(lca(ask[i].l,ask[i].r));        int l=0,r=n,ans=0;        while(l<=r){            int mid=(l+r)>>1;            if(check(mid))ans=mid+1,l=mid+1;            else r=mid-1;        }        Ans[ask[i].id]=ans;        reverse(lca(ask[i].l,ask[i].r));    }    for(int i=1;i<=cnt1;i++)printf("%d\n",Ans[i]);}