题解 超级加倍
链上的部分分可以单调栈求出范围 \(l, r\) 后主席树维护
其实也可以求出后转化为三维偏序求解
题解说可以忽略一个条件，再减去算重的
然后正解
- 与形如 经过点中的最大值/起点为全路径最大值 类似的问题，序列上可以考虑笛卡尔树，树上可以考虑kruskal重构树
- 对点权建立kruskal重构树（lca为最大值）：
从小到大扫描点，扫描当前点的所有出边 \((u, v)\)，仅当 \(v<u\) 时在 \(u\) 和 \(find(v)\) 间连一条边
于是可以依据点权建两棵kruskal重构树
现在问题变为了求 \(x\) 在 \(T1\) 中是 y 的祖先，\(y\) 在 \(T2\) 中是 \(x\) 的祖先的点对 \((x, y)\) 数量
是个二维偏序问题，可以求出 \(T1\) 的dfs序，在 \(T2\) 上dfs
dfs时保证当前dfs到的点的每个祖先在其在 \(T1\) dfs序上的位置产生1的贡献
于是可以树状数组查询当前点子树内1的个数
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 2000010
#define ll long long
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n;
int head[N], dep[N], mdep[N], deg[N], a[N], tot, size;
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++size].to=t; e[size].next=head[s]; head[s]=size;}
void dfs1(int u, int fa) {
mdep[u]=dep[u];
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v!=fa) {
dep[v]=dep[u]+1;
dfs1(v, u);
mdep[u]=max(mdep[u], mdep[v]);
}
}
}
void dfs3(int u, int fa) {
a[++tot]=u;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v!=fa) dfs3(v, u);
}
}
namespace force{
int ans;
void dfs2(int u, int rot, int fa, int maxn) {
// cout<<"dfs: "<<u<<' '<<rot<<' '<<maxn<<endl;
if (maxn<u) ++ans; //, cout<<rot<<' '<<u<<endl;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v!=fa && v>rot) dfs2(v, rot, u, max(maxn, u));
}
}
void solve() {
for (int i=1; i<=n; ++i) dfs2(i, i, 0, i);
printf("%d\n", ans);
exit(0);
}
}
namespace task1{
ll ans;
void solve() {
int rot;
for (int i=1; i<=n; ++i) if (deg[i]==n-1) {rot=i; break;}
for (int i=1; i<=n; ++i) if (i!=rot) {
++ans;
if (i<rot) ans+=n-rot;
}
printf("%lld\n", ans);
exit(0);
}
}
namespace task2{
ll ans;
int l[N], r[N];
int q[N], ql, qr;
int ls[N*10], rs[N*10], sum[N*10], rot[N], now;
#define pushup(p) sum[p]=sum[ls[p]]+sum[rs[p]]
void upd(int& p1, int p2, int tl, int tr, int pos, int val) {
if (!p1) {p1=++now;}
if (tl==tr) {sum[p1]=sum[p2]+val; return ;}
int mid=(tl+tr)>>1;
if (pos<=mid) upd(ls[p1], ls[p2], tl, mid, pos, val), rs[p1]=rs[p2];
else upd(rs[p1], rs[p2], mid+1, tr, pos, val), ls[p1]=ls[p2];
pushup(p1);
}
int query(int p1, int p2, int tl, int tr, int ql, int qr) {
if (!p1) return 0;
if (ql<=tl && qr>=tr) {return sum[p2]-sum[p1];}
int mid=(tl+tr)>>1, ans=0;
if (ql<=mid) ans+=query(ls[p1], ls[p2], tl, mid, ql, qr);
if (qr>mid) ans+=query(rs[p1], rs[p2], mid+1, tr, ql, qr);
return ans;
}
void solve() {
ql=1; qr=0; a[n+1]=0;
for (int i=1; i<=n+1; ++i) {
while (ql<=qr && a[q[qr]]>a[i]) r[q[qr--]]=i-1;
q[++qr]=i;
}
ql=1; qr=0; a[0]=n+1;
for (int i=n; ~i; --i) {
while (ql<=qr && a[q[qr]]<a[i]) l[q[qr--]]=i+1;
q[++qr]=i;
}
// cout<<"l: "; for (int i=1; i<=n; ++i) cout<<l[i]<<' '; cout<<endl;
// cout<<"r: "; for (int i=1; i<=n; ++i) cout<<r[i]<<' '; cout<<endl;
for (int i=1; i<=n; ++i) upd(rot[i], rot[i-1], 1, n, l[i], 1);
for (int i=1; i<=n; ++i) if (r[i]>i) ans+=query(rot[i], rot[r[i]], 1, n, 1, i);
now=0;
memset(l, 0, sizeof(l));
memset(r, 0, sizeof(r));
memset(ls, 0, sizeof(ls));
memset(rs, 0, sizeof(rs));
memset(sum, 0, sizeof(sum));
memset(rot, 0, sizeof(rot));
reverse(a+1, a+n+1);
ql=1; qr=0; a[n+1]=0;
for (int i=1; i<=n+1; ++i) {
while (ql<=qr && a[q[qr]]>a[i]) r[q[qr--]]=i-1;
q[++qr]=i;
}
ql=1; qr=0; a[0]=n+1;
for (int i=n; ~i; --i) {
while (ql<=qr && a[q[qr]]<a[i]) l[q[qr--]]=i+1;
q[++qr]=i;
}
// cout<<"l: "; for (int i=1; i<=n; ++i) cout<<l[i]<<' '; cout<<endl;
// cout<<"r: "; for (int i=1; i<=n; ++i) cout<<r[i]<<' '; cout<<endl;
for (int i=1; i<=n; ++i) upd(rot[i], rot[i-1], 1, n, l[i], 1);
for (int i=1; i<=n; ++i) if (r[i]>i) ans+=query(rot[i], rot[r[i]], 1, n, 1, i);
printf("%lld\n", ans);
exit(0);
}
}
namespace task{
ll ans;
int dsu[N], id[N], siz[N], bit[N], now;
inline void upd(int i, int val) {for (; i<=n; i+=i&-i) bit[i]+=val;}
inline int query(int i) {int ans=0; for (; i; i-=i&-i) ans+=bit[i]; return ans;}
inline int find(int p) {return dsu[p]==p?p:dsu[p]=find(dsu[p]);}
namespace tr1{
int head[N], size;
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++size].to=t; e[size].next=head[s]; head[s]=size;}
}
namespace tr2{
int head[N], size;
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++size].to=t; e[size].next=head[s]; head[s]=size;}
}
void dfs1(int u) {
using namespace tr1;
siz[u]=1;
id[u]=++now;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
dfs1(v);
siz[u]+=siz[v];
}
}
void dfs2(int u) {
using namespace tr2;
ans+=query(id[u]+siz[u]-1)-query(id[u]);
upd(id[u], 1);
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
dfs2(v);
}
upd(id[u], -1);
}
void solve() {
memset(tr1::head, -1, sizeof(tr1::head));
memset(tr2::head, -1, sizeof(tr2::head));
for (int i=1; i<=n; ++i) dsu[i]=i;
for (int i=1; i<=n; ++i) {
for (int j=head[i],v; ~j; j=e[j].next) {
v = e[j].to;
if (v<i && find(v)!=find(i)) {
tr1::add(i, find(v));
dsu[find(v)]=i;
}
}
}
dfs1(n);
for (int i=1; i<=n; ++i) dsu[i]=i;
for (int i=n; i; --i) {
for (int j=head[i],v; ~j; j=e[j].next) {
v = e[j].to;
if (v>i && find(v)!=find(i)) {
tr2::add(i, find(v));
dsu[find(v)]=i;
}
}
}
dfs2(1);
printf("%lld\n", ans);
exit(0);
}
}
signed main()
{
freopen("charity.in", "r", stdin);
freopen("charity.out", "w", stdout);
n=read(); read();
memset(head, -1, sizeof(head));
for (int i=2,u; i<=n; ++i) {
u=read();
add(u, i); add(i, u);
++deg[u]; ++deg[i];
}
#if 0
if (n<=1000) force::solve();
dep[1]=1; dfs1(1, 0);
int odd=0, beg;
for (int i=1; i<=n; ++i) {
if (deg[i]==1) ++odd, beg=i;
else if (deg[i]>2) odd=INF;
}
if (odd==2) {dfs3(beg, 0); task2::solve();}
int mdeg=0;
for (int i=1; i<=n; ++i) mdeg=max(mdeg, deg[i]);
if (mdeg==n-1) task1::solve();
force::solve();
#else
task::solve();
#endif
return 0;
}
