标签：int 题解 top Revive dfs2 mson ans mod
柿子人畜无害，但有个地方误导性极强
- 给定一棵树，每条边有一个边权，要带修查询一个点与其子树外的所有点间的距离和
这个东西没有可以时间可以接受的解法!
考场上就死这了……觉得维护出来就可以A了，结果死活维护不出来
正解是另一种解法：
- \((\sum a_i)^2 = \sum a_i^2 + \sum\limits_{x<y} 2*a_x*a_y\)
首先初始 \(ans\) 可以很方便地树形DP求出来，维护 \(\sum\limits_{v\in u's\ tree}dis(u, v)\) 和 \(\sum\limits_{v\in u's\ tree}dis^2(u, v)\) ，考虑路径合并就行了
于是考虑如何维护同时经过 \(a, b\) 两条边的路径数
\(a\) 是给定的，\(b\) 要分情况讨论：在 \(a\) 的子树里，在 \(a\) 到根节点的路径上，在其它子树中
那可以拆成dfs序用树状数组维护
case 1：直接维护
case 2：差分一下也可以直接维护
case 3：开两个树状数组，把贡献打在离开子树时同样可以维护
复杂度 \(O(n+(n+q)*13logn)\)，完全不卡常跑得巨快，吊打线段树 \(O(n+q*2logn)\) 复杂度
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#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, q, sub;
int head[N], size, dep[N], id[N], rk[N], tot=1, top[N], fa[N], msiz[N], mson[N];
ll val[N], sum[N], siz[N];
const ll mod=1e9+7;
inline void md(ll& a, ll b) {a+=b; a=a>=mod?a-mod:a;}
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{
ll bin[N];
inline void upd(int i, ll dat) {for (; i<=tot; i+=i&-i) md(bin[i], dat);}
inline ll query(int i) {ll ans=0; for (; i; i-=i&-i) md(ans, bin[i]); return ans;}
void build() {for (int i=2; i<=n; ++i) upd(id[i], val[i]%mod);}
void dfs1(int u, int pa) {
siz[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==pa) continue;
dep[v]=dep[u]+1, fa[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 f, int t) {
top[u]=t;
id[u]=++tot;
rk[tot]=u;
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==f || v==mson[u]) continue;
dfs2(v, u, v);
}
}
ll qsum(int a, int b) {
ll ans=0;
while (top[a]!=top[b]) {
if (dep[top[a]]<dep[top[b]]) swap(a, b);
ans=(ans+query(id[a])-query(id[top[a]]-1))%mod;
a=fa[top[a]];
}
if (dep[a]>dep[b]) swap(a, b);
ans=(ans+query(id[b])-query(id[a]))%mod;
return (ans+mod)%mod;
}
void solve() {
dep[1]=1; dfs1(1, 0); dfs2(1, 0, 1); build();
ll ans=0, tem;
for (int i=1; i<=n; ++i)
for (int j=i+1; j<=n; ++j) {
tem=qsum(i, j);
ans=(ans+tem*tem)%mod;
}
printf("%lld\n", ans);
for (int i=1,u,w; i<=q; ++i) {
u=read(); w=read();
upd(id[u], w);
ans=0;
for (int j=1; j<=n; ++j)
for (int k=j+1; k<=n; ++k) {
tem=qsum(j, k);
ans=(ans+tem*tem)%mod;
}
printf("%lld\n", ans);
}
exit(0);
}
}
namespace task1{
ll bin[N], sva[N], sva2[N], ans;
inline void upd(int i, ll dat) {for (; i<=tot; i+=i&-i) md(bin[i], dat);}
inline ll query(int i) {ll ans=0; for (; i; i-=i&-i) md(ans, bin[i]); return ans;}
void build() {for (int i=2; i<=n; ++i) upd(id[i], val[i]%mod);}
void dfs1(int u, int pa) {
siz[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==pa) continue;
dep[v]=dep[u]+1, fa[v]=u, dfs1(v, u);
ans = (ans + (siz[v]*sva2[u])%mod
+ (siz[u]-1)*((sva2[v]+2*val[v]*sva[v]%mod+siz[v]*val[v]%mod*val[v]%mod)%mod)%mod
+ 2*sva[u]*(sva[v]+siz[v]*val[v]%mod)%mod)%mod;
siz[u]+=siz[v];
sva2[u]=(sva2[u]+sva2[v]+2*val[v]*sva[v]%mod+siz[v]*val[v]%mod*val[v]%mod)%mod;
sva[u]=(sva[u]+sva[v]+siz[v]*val[v])%mod;
if (siz[v]>msiz[u]) msiz[u]=siz[v], mson[u]=v;
}
ans=(ans+sva2[u])%mod;
}
void dfs2(int u, int f, int t) {
top[u]=t;
id[u]=++tot;
rk[tot]=u;
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==f || v==mson[u]) continue;
dfs2(v, u, v);
}
}
ll qsum(int a, int b) {
ll ans=0;
while (top[a]!=top[b]) {
if (dep[top[a]]<dep[top[b]]) swap(a, b);
ans=(ans+query(id[a])-query(id[top[a]]-1))%mod;
a=fa[top[a]];
}
if (dep[a]>dep[b]) swap(a, b);
ans=(ans+query(id[b])-query(id[a]))%mod;
return (ans+mod)%mod;
}
void solve() {
dep[1]=1; dfs1(1, 0); dfs2(1, 0, 1); build();
ll tem, dlt;
printf("%lld\n", ans);
for (int i=1,u,w; i<=q; ++i) {
u=read(); w=read(); tem=0, dlt=0;
for (int now=fa[u],lst=u; now; lst=now,now=fa[now]) {
dlt = (dlt+val[lst])%mod;
tem = (tem + sva[now]-sva[lst]-val[lst]*siz[lst]%mod + (siz[now]-siz[lst])*dlt%mod)%mod;
}
tem=(tem*siz[u]%mod + sva[u]*(n-siz[u]))%mod;
ans = (ans+2ll*w*tem%mod+(n-siz[u])*siz[u]%mod*w%mod*w%mod)%mod;
val[u]=(val[u]+w)%mod;
for (int now=fa[u]; now; now=fa[now]) {
sva[now]=(sva[now]+1ll*w*siz[u]%mod)%mod;
}
printf("%lld\n", (ans%mod+mod)%mod);
}
exit(0);
}
}
namespace task{
ll in_tree[N<<1], on_chain[N<<1], other_tree[N<<1], other_dlt[N<<1], sva[N], sva2[N], ans, in[N], out[N];
inline void upd(ll* bin, int i, ll dat) {for (; i<=tot; i+=i&-i) bin[i]=(bin[i]+dat)%mod;}
inline ll query(ll* bin, int i) {ll ans=0; for (; i; i-=i&-i) ans=(ans+bin[i])%mod; return ans;}
void build() {
for (int i=2; i<=n; ++i) {
//cout<<"i: "<<i<<' '<<in[i]<<' '<<out[i]<<endl;
upd(in_tree, in[i], val[i]*siz[i]%mod);
upd(on_chain, in[i], val[i]*(n-siz[i])%mod);
upd(on_chain, out[i], -val[i]*(n-siz[i])%mod);
upd(other_dlt, in[i], -val[i]*siz[i]%mod);
upd(other_dlt, out[i], val[i]*siz[i]%mod);
upd(other_tree, out[i], val[i]*siz[i]%mod);
}
}
void dfs1(int u, int pa) {
//cout<<"dfs1 "<<u<<' '<<pa<<endl;
siz[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==pa) continue;
dep[v]=dep[u]+1, fa[v]=u, dfs1(v, u);
ans = (ans + (siz[v]*sva2[u])%mod
+ (siz[u]-1)*((sva2[v]+2*val[v]*sva[v]%mod+siz[v]*val[v]%mod*val[v]%mod)%mod)%mod
+ 2*sva[u]*(sva[v]+siz[v]*val[v]%mod)%mod)%mod;
siz[u]+=siz[v];
sva2[u]=(sva2[u]+sva2[v]+2*val[v]*sva[v]%mod+siz[v]*val[v]%mod*val[v]%mod)%mod;
sva[u]=(sva[u]+sva[v]+siz[v]*val[v]%mod)%mod;
if (siz[v]>msiz[u]) msiz[u]=siz[v], mson[u]=v;
}
ans=(ans+sva2[u])%mod;
}
void dfs2(int u, int f, int t) {
top[u]=t;
in[u]=id[u]=++tot;
rk[tot]=u;
if (!mson[u]) {out[u]=++tot; return ;}
dfs2(mson[u], u, t);
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (v==f || v==mson[u]) continue;
dfs2(v, u, v);
}
out[u]=++tot;
}
ll qsum(int a, int b) {
ll ans=0;
while (top[a]!=top[b]) {
if (dep[top[a]]<dep[top[b]]) swap(a, b);
ans=(ans+query(on_chain, id[a])-query(on_chain, id[top[a]]-1))%mod;
a=fa[top[a]];
}
if (dep[a]>dep[b]) swap(a, b);
ans=(ans+query(on_chain, id[b])-query(on_chain, id[a]))%mod;
return (ans+mod)%mod;
}
void solve() {
dep[1]=1; dfs1(1, 0); dfs2(1, 0, 1); build();
ll w;
printf("%lld\n", ans);
for (int i=1,u; i<=q; ++i) {
//cout<<"i: "<<i<<endl;
u=read(); w=read();
ans = (ans+siz[u]*(n-siz[u])%mod*(2*val[u]*w%mod+w*w%mod)%mod)%mod;
#if 0
cout<<"ans1: "<<ans<<endl;
cout<<"query: "<<2*w*query(other_tree, in[u])<<' '<<
2*w*query(other_dlt, in[u]-1)<<' '<<2*w*(query(other_tree, tot)-query(other_tree, out[u]))<<endl;
cout<<"assert: "<<query(other_tree, out[2])<<endl;
cout<<"now: "<< (n-siz[u])*(query(in_tree, out[u])-query(in_tree, in[u]))%mod<<' '<<
siz[u]*query(on_chain, in[u]-1)%mod<<' '<<
siz[u]*((query(other_tree, in[u])+
query(other_dlt, in[u]-1)+query(other_tree, tot)-query(other_tree, out[u]))%mod)%mod<<endl;
#endif
ans = (ans + 2*w*( (n-siz[u])*(query(in_tree, out[u])-query(in_tree, in[u]))%mod
+ siz[u]*query(on_chain, in[u]-1)%mod
+ siz[u]*((query(other_tree, in[u])
+query(other_dlt, in[u]-1)+query(other_tree, tot)-query(other_tree, out[u]))%mod)%mod)%mod)%mod;
val[u]=(val[u]+w)%mod;
upd(in_tree, in[u], w*siz[u]%mod);
upd(on_chain, in[u], w*(n-siz[u])%mod);
upd(on_chain, out[u], -w*(n-siz[u])%mod);
upd(other_dlt, in[u], -w*siz[u]%mod);
upd(other_dlt, out[u], w*siz[u]%mod);
upd(other_tree, out[u], w*siz[u]%mod);
printf("%lld\n", (ans%mod+mod)%mod);
}
exit(0);
}
}
signed main()
{
memset(head, -1, sizeof(head));
sub=read(); n=read(); q=read();
for (int i=2,u; i<=n; ++i) {
u=read(); val[i]=read();
add(u, i); add(i, u);
}
task::solve();
return 0;
}
标签：int,题解,top,Revive,dfs2,mson,ans,mod 来源： https://www.cnblogs.com/narration/p/15195233.html
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