BZOJ 4771: 七彩树 可持久化线段树+树链的并

这个思路挺有意思的 ~

利用树链的并来保证每个颜色只贡献一次,然后用可持久化线段树维护

code:

#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 100005
#define setIO(s) freopen(s".in","r",stdin) , freopen(s".out","w",stdout)
using namespace std;
namespace seg
{
#define lson t[x].ls
#define rson t[x].rs
int tot;
struct node
{
int ls,rs,sum;
}t[N*80];
void clr()
{
for(int i=1;i<=tot;++i) t[i].ls=t[i].rs=t[i].sum=0;
tot=0;
}
int newnode()
{
return ++tot;
}
void build(int &x,int l,int r)
{
x=newnode();
if(l==r) return;
int mid=(l+r)>>1;
if(l<=mid) build(lson,l,mid);
if(r>mid) build(rson,mid+1,r);
}
int cop(int x,int l,int r,int p,int v)
{
int now=newnode();
t[now]=t[x];
t[now].sum+=v;
if(l==r) return now;
int mid=(l+r)>>1;
if(p<=mid) t[now].ls=cop(lson,l,mid,p,v);
else t[now].rs=cop(rson,mid+1,r,p,v);
return now;
}
int query(int x,int l,int r,int L,int R)
{
if(!x) return 0;
if(l>=L&&r<=R) return t[x].sum;
int re=0;
int mid=(l+r)>>1;
if(L<=mid) re+=query(lson,l,mid,L,R);
if(R>mid) re+=query(rson,mid+1,r,L,R);
return re;
}
#undef lson
#undef rson
};
set<int>se[N];
set<int>::iterator fr,ba;
int edges,n,m,tim,ct;
int hd[N],to[N],nex[N],col[N],nod[N],st[N],ed[N];
int fa[N],dfn[N],dep[N],size[N],son[N],top[N];
int id[N],rt[N];
bool cmp(int a,int b)
{
return dep[a]<dep[b];
}
void add(int u,int v)
{
nex[++edges]=hd[u],hd[u]=edges,to[edges]=v;
}
void dfs1(int u,int ff)
{
size[u]=1;
dfn[u]=++tim;
st[u]=dfn[u];
nod[dfn[u]]=u;
dep[u]=dep[ff]+1;
for(int i=hd[u];i;i=nex[i])
{
int v=to[i];
dfs1(v,u);
size[u]+=size[v];
if(size[v]>size[son[u]]) son[u]=v;
}
ed[u]=tim;
}
void dfs2(int u,int tp)
{
top[u]=tp;
if(son[u]) dfs2(son[u],tp);
for(int i=hd[u];i;i=nex[i])
{
if(to[i]!=son[u])
{
dfs2(to[i],to[i]);
}
}
}
int LCA(int x,int y)
{
while(top[x]!=top[y])
{
dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];
}
return dep[x]<dep[y]?x:y;
}
void solve()
{
int i,j,last=0;
scanf("%d%d",&n,&m);
for(i=1;i<=n;++i) scanf("%d",&col[i]);
for(i=2;i<=n;++i)
{
scanf("%d",&fa[i]);
add(fa[i],i);
}
dfs1(1,0);
dfs2(1,1);
for(i=1;i<=n;++i) id[i]=i;
sort(id+1,id+1+n,cmp);
int lst=0;
seg::build(rt[0],1,n);
for(i=1;i<=n;++i)
{
int p=id[i];
int c=col[p];
while(dep[p]>lst) rt[lst+1]=rt[lst],++lst;
se[c].insert(dfn[p]);
fr=ba=se[c].lower_bound(dfn[p]),ba++;
if(fr!=se[c].begin())
{
--fr;
rt[lst]=seg::cop(rt[lst],1,n,dfn[LCA(nod[*fr],p)],-1);
++fr;
}
if(ba!=se[c].end())
{
rt[lst]=seg::cop(rt[lst],1,n,dfn[LCA(nod[*ba],p)],-1);
}
if(fr!=se[c].begin()&&ba!=se[c].end())
{
--fr;
rt[lst]=seg::cop(rt[lst],1,n,dfn[LCA(nod[*fr],nod[*ba])],1);
}
rt[lst]=seg::cop(rt[lst],1,n,dfn[p],1);
}
for(i=1;i<=m;++i)
{
int x,d;
scanf("%d%d",&x,&d);
x^=last,d^=last;
last=seg::query(rt[min(n,dep[x]+d)],1,n,st[x],ed[x]);
printf("%d\n",last);
}
for(i=1;i<=n;++i) son[i]=0;
for(i=1;i<=n;++i) hd[i]=0;
for(i=1;i<=edges;++i) nex[i]=to[i]=0;
edges=tim=0;
seg::clr();
for(i=1;i<=n;++i) se[i].clear();
}
int main()
{
// setIO("input");
int i,j,T;
scanf("%d",&T);
while(T--) solve();
return 0;
}

  

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