Codeforces696E

Codeforces696E题解

题解：

• 一个一个删，树剖线段树上维护即可。

代码：

#include <bits/stdc++.h>
#define gc getchar()
#define ll long long
#define mid (l+r>>1)
#define inf (ll)2e18
#define N 100009
#define root 1,1,n
#define lc cur<<1
#define rc lc|1
#define lson lc,l,mid
#define rson rc,mid+1,r
#define now cur,l,r
using namespace std;
ll n,m,Q,first[N],number,fa[N],size[N],top[N],deep[N],Mson[N],dfn[N],id[N],ed[N],cnt;
ll tg[N<<2],print[N],pos[N];
struct edge
{
    ll to,next;
    void add(ll x,ll y)
    {
        to=y,next=first[x],first[x]=number;
    }
}e[N<<1];
struct node
{
    ll pos,val,id;
    node(ll pos=0,ll val=0,ll id=0):pos(pos),val(val),id(id){}
    bool operator <(const node &rhs) const
    {
        return val<rhs.val||(val==rhs.val&&pos<rhs.pos);
    }
}Min[N<<2];
vector<node> a[N];
ll read()
{
	ll x=1;
	char ch;
	while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;
	ll s=ch-'0';
	while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';
	return s*x;
}
void dfs1(ll x)
{
    deep[x]=deep[fa[x]]+1;
    size[x]=1;
    for (ll i=first[x];i;i=e[i].next)
        if (e[i].to!=fa[x])
        {
            fa[e[i].to]=x;
            dfs1(e[i].to);
            size[x]+=size[e[i].to];
            if (size[e[i].to]>size[Mson[x]]) Mson[x]=e[i].to;
        }
}
void dfs2(ll x,ll y)
{
    top[x]=y;
    id[dfn[x]=++cnt]=x;
    if (Mson[x]) dfs2(Mson[x],y);
    for (ll i=first[x];i;i=e[i].next)
        if (e[i].to!=fa[x]&&e[i].to!=Mson[x]) dfs2(e[i].to,e[i].to);
    ed[x]=cnt;
}
void down(ll cur,ll l,ll r)
{
    if (tg[cur]>0)
    {
        tg[lc]+=tg[cur];
        tg[rc]+=tg[cur];
        Min[lc].val+=tg[cur];
        Min[rc].val+=tg[cur];
        tg[cur]=0;
    }
}
void up(ll cur,ll l,ll r)
{
    Min[cur]=min(Min[lc],Min[rc]);
}
void build(ll cur,ll l,ll r)
{
    tg[cur]=0;
    if (l==r)
    {
        Min[cur]=(a[id[l]].size()!=0?a[id[l]][0]:node(0,inf,inf));
        return;
    }
    build(lson);
    build(rson);
    up(now);
}
void ins(ll cur,ll l,ll r,ll L,ll R,ll x)
{
    if (L<=l&&R>=r)
    {
        tg[cur]+=x;
        Min[cur].val+=x;
        return;
    }
    down(now);
    if (L<=mid) ins(lson,L,R,x);
    if (R>mid) ins(rson,L,R,x);
    up(now);
}
ll get_tg(ll cur,ll l,ll r,ll x)
{
    if (l==r) return tg[cur];
    if (x<=mid) return tg[cur]+get_tg(lson,x);
    else return tg[cur]+get_tg(rson,x);
}
void del(ll cur,ll l,ll r,ll x)
{
    if (l==r)
    {
        pos[id[l]]++;
        Min[cur]=(a[id[l]].size()>pos[id[l]]?a[id[l]][pos[id[l]]]:node(0,inf,inf));
        if (a[id[l]].size()>pos[id[l]]) Min[cur].val+=get_tg(root,l);
        return;
    }
    down(now);
    if (x<=mid) del(lson,x);
    else del(rson,x);
    up(now);
}
node qry(ll cur,ll l,ll r,ll L,ll R)
{
    if (L<=l&&R>=r) return Min[cur];
    down(now);
    node ret=node(0,inf,inf);
    if (L<=mid) ret=min(ret,qry(lson,L,R));
    if (R>mid) ret=min(ret,qry(rson,L,R));
    return ret;
}
node qry(ll x,ll y)
{
    node ret=node(0,inf,inf);
    for (;top[x]!=top[y];x=fa[top[x]])
    {
        if (deep[top[x]]<deep[top[y]]) swap(x,y);
        ret=min(ret,qry(root,dfn[top[x]],dfn[x]));
    }
    if (dfn[x]>dfn[y]) swap(x,y);
    return min(ret,qry(root,dfn[x],dfn[y]));
}
int main()
{
    n=read(),m=read(),Q=read();
    //n=1,m=100000,Q=100000;
    for (ll i=1;i<n;i++)
    {
        ll x=read(),y=read();
        e[++number].add(x,y),e[++number].add(y,x);
    }
    for (ll i=1;i<=m;i++)
    {
        ll x=read();
        //ll x=1;
        a[x].push_back(node(x,i,i));
    }
    for (ll i=1;i<=n;i++) pos[i]=0,sort(a[i].begin(),a[i].end());
    dfs1(1);
    dfs2(1,1);
    build(root);
    while (Q--)
    {
        ll op=read();
        //ll op=1;
        if (op==1)
        {
            ll x=read(),y=read(),k=read(),num=0;
            //ll x=1,y=1,k=1,num=0;
            for (num=0;num<k;num++)
            {
                node Now=qry(x,y);
                if (Now.pos==0) break;
                else
                {
                    print[num+1]=Now.id;
                    del(root,dfn[Now.pos]);
                }
            }
            printf("%lld",num);
            for (ll i=1;i<=num;i++) printf(" %lld",print[i]);
            puts("");
        }
        else
        {
            ll x=read(),k=read();
            ins(root,dfn[x],ed[x],k);
        }
    }
	return 0;
}