题解 树上排列
并不会做
原题在这里
- 对于两个无序序列是否相等的判断：可以用无序hash
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 500010
#define ll long long
#define ull unsigned 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, q;
int a[N];
int head[N], size;
ull pw[N], pre[N];
const ull base=1e9+7;
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 force{
int sta[N], cnt[N], top;
int dfs(int u, int fa, int to) {
if (u==to) {sta[++top]=a[u]; return 1;}
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v!=fa) {
int t=dfs(v, u, to);
if (t) {sta[++top]=a[u]; return 1;}
}
}
return 0;
}
void solve() {
for (int i=1,x,y; i<=q; ++i) {
if (read()&1) {
x=read(); y=read(); top=0;
dfs(x, 0, y);
for (int i=1; i<=top; ++i) cnt[i]=0;
for (int i=1; i<=top; ++i) ++cnt[sta[i]];
for (int i=1; i<=top; ++i) if (cnt[i]!=1) {puts("No"); goto jump;}
puts("Yes");
jump: ;
}
else {
x=read(); y=read();
a[x]=y;
}
}
exit(0);
}
}
namespace task{
int top[N], dep[N], siz[N], msiz[N], mson[N], id[N], rk[N], back[N], tot;
int tl[N<<2], tr[N<<2];
ull sum[N<<2];
#define tl(p) tl[p]
#define tr(p) tr[p]
#define sum(p) sum[p]
#define pushup(p) sum(p)=sum(p<<1)+sum(p<<1|1)
void build(int p, int l, int r) {
tl(p)=l; tr(p)=r;
if (l==r) {sum(p)=pw[a[rk[l]]]; return ;}
int mid=(l+r)>>1;
build(p<<1, l, mid);
build(p<<1|1, mid+1, r);
pushup(p);
}
void upd(int p, int pos, int val) {
if (tl(p)==tr(p)) {sum(p)=pw[val]; return ;}
int mid=(tl(p)+tr(p))>>1;
if (pos<=mid) upd(p<<1, pos, val);
else upd(p<<1|1, pos, val);
pushup(p);
}
ull query(int p, int l, int r) {
if (l<=tl(p)&&r>=tr(p)) return sum(p);
int mid=(tl(p)+tr(p))>>1; ull ans=0;
if (l<=mid) ans+=query(p<<1, l, r);
if (r>mid) ans+=query(p<<1|1, l, r);
return ans;
}
void dfs1(int u, int fa) {
siz[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa) continue;
dep[v]=dep[u]+1; back[v]=u;
dfs1(v, u);
siz[u]+=siz[v];
if (siz[v]>msiz[u]) msiz[u]=siz[v], mson[u]=v;
}
}
void dfs2(int u, int fa, int t) {
id[u]=++tot;
rk[tot]=u;
top[u]=t;
if (!mson[u]) return ;
dfs2(mson[u], u, t);
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==fa || v==mson[u]) continue;
dfs2(v, u, v);
}
}
int lca(int a, int b) {
while (top[a]!=top[b]) {
if (dep[top[a]]<dep[top[b]]) swap(a, b);
a=back[top[a]];
}
if (dep[a]>dep[b]) swap(a, b);
return a;
}
inline int dis(int a, int b) {return dep[a]+dep[b]-2*dep[lca(a, b)]+1;}
ll qsum(int a, int b) {
ll ans=0;
while (top[a]!=top[b]) {
if (dep[top[a]]<dep[top[b]]) swap(a, b);
// cout<<"id: "<<top[a]<<' '<<a<<endl;
ans+=query(1, id[top[a]], id[a]);
a=back[top[a]];
}
if (dep[a]>dep[b]) swap(a, b);
// cout<<"q: "<<a<<' '<<b<<endl;
ans+=query(1, id[a], id[b]);
return ans;
}
void solve() {
tot=0;
memset(msiz, 0, sizeof(msiz));
memset(mson, 0, sizeof(mson));
dep[1]=1; dfs1(1, 0); dfs2(1, 0, 1);
build(1, 1, n);
for (int i=1,x,y; i<=q; ++i) {
if (read()&1) {
x=read(); y=read();
// if (x==y) {puts("Yes"); continue;}
// cout<<qsum(x, y)<<endl;
// cout<<"dis: "<<dis(x, y)<<endl;
// cout<<"pre: "<<pre[dis(x, y)]<<endl;
puts(qsum(x, y)==pre[dis(x, y)]?"Yes":"No");
}
else {
x=read(); y=read();
upd(1, id[x], y);
}
}
}
}
signed main()
{
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
int T=read();
pw[0]=1;
for (int i=1; i<N; ++i) pw[i]=pw[i-1]*base;
for (int i=1; i<N; ++i) pre[i]=pre[i-1]+pw[i];
while (T--) {
n=read(); q=read();
memset(head, -1, sizeof(head)); size=0;
for (int i=1; i<=n; ++i) a[i]=read();
for (int i=1,u,v; i<n; ++i) {
u=read(); v=read();
add(u, v); add(v, u);
}
// force::solve();
task::solve();
}
return 0;
}
