[点分树] BZOJ 3924 [Zjoi2015]幻想乡战略游戏

%%%CLJ

首先动态维护每个点的答案？动态树分治就好了

怎么找重心

对于分治结构的每一个点，我们枚举它的出边

如果某条出边连向的点的距离之和小于当前点，那么答案一定在那条出边指向的子树中，分治做下去就行了
这样是两个log再乘20的  实际上最优可以一个log

跑的竟然还没有当年不懂事打的暴力快

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=200005;
const int K=25;

struct edge{
int u,v,w,next;
}G[N<<1];
int head1
,head2
,inum;
int *head;

inline void add(int u,int v,int w,int p,int *head=::head){
G[p].u=u; G[p].v=v; G[p].w=w; G[p].next=head[u]; head[u]=p;
}

#define V G[p].v

int size
,del
;
int sum,rt,minimum;

int fat
[K],dis
[K];

inline void Root(int u,int fa) {
int maximum=0;
size[u]=1;
for (int p=head[u];p;p=G[p].next)
if (V!=fa && !del[V]) {
Root(V,u);
size[u]+=size[V];
maximum=max(maximum,size[V]);
}
maximum=max(maximum,sum-size[u]);
if (minimum>maximum) minimum=maximum,rt=u;
}

inline void Depth(int u,int fa,int d,int f){
size[u]=1; fat[u][++*fat[u]]=f; dis[u][++*dis[u]]=d;
for (int p=head[u];p;p=G[p].next)
if (!del[V] && V!=fa)
Depth(V,u,d+G[p].w,f),size[u]+=size[V];
}

int fad
;

inline void Divi(int u) {
del[u]=1;
Depth(u,0,0,u);
for (int p=head[u];p;p=G[p].next)
if (!del[V]){
sum=size[V]; minimum=1<<30; Root(V,0);
add(u,rt,0,++inum,head2); fad[rt]=V;
Divi(rt);
}
}

int n,Q,root;
ll tot
,ans
,fav
,fatot
,Tot;

inline ll Calc(int t){
ll ret=ans[t];
for (int j=2;j<=*fat[t];j++)
ret+=(ans[fat[t][j]]-fav[fat[t][j-1]])+(tot[fat[t][j]]-fatot[fat[t][j-1]])*dis[t][j];
return ret;
}

int main(){
int iu,iv,iw;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); read(Q); head=head1;
for (int i=1;i<n;i++)
read(iu),read(iv),read(iw),add(iv,iu,iw,++inum),add(iu,iv,iw,++inum);
sum=n; minimum=1<<30; Root(1,0);
root=rt; Divi(rt);
for (int i=1;i<=n;i++) reverse(fat[i]+1,fat[i]+*fat[i]+1),reverse(dis[i]+1,dis[i]+*dis[i]+1);
head=head2;
while (Q--){
read(iu); read(iv);
Tot+=iv;
for (int j=1;j<=*fat[iu];j++){
ans[fat[iu][j]]+=(ll)iv*dis[iu][j],tot[fat[iu][j]]+=iv;
fav[fat[iu][j]]+=(ll)iv*dis[iu][j+1]; fatot[fat[iu][j]]+=iv;
}
int t=root;
while (1){
int flag=0;
for (int p=head[t];p;p=G[p].next)
if (Calc(fad[V])<Calc(t)){
t=V; flag=1; break;
}
if (!flag) break;
}
printf("%lld\n",Calc(t));
}
return 0;
}