9-17考试 数星星
题意
给定一棵树，点带权，m个点对（m条路径），q次询问区间路径并的点权和。
\(n,m,q\le 10^5\) ### 题解
考虑移动右端点。
对每个点维护最后一个包含它的路径编号\(x_i\)。
询问就相当于\(x_i\ge l\)的点权和。
操作就相当于链推平。
树剖之后相当于链推平。
用一个类似于珂朵莉树的东西维护连续段就可以了
具体的，用set<pair<int,int> int>来维护连续段和它的\(x\)，每次推平l,r就相当于在l-1处拆分，r处拆分，然后把l..r的区间全部删掉，最后加一个l,r的区间就可以了
删区间加区间的时候可以用树状数组维护点权和。
时间复杂度\(O(m\log^2 n)\)，为啥？
每次推平最多拆两次（多两个区间），每个区间最多被删一次， 每次最多加一个区间，所以均摊\(O(log(n))\)一次推平（因为要用set定位），又因为要树剖所以就\(O(m\log n)\)次推平。
代码
/*
Author: QAQ Automaton
Lang: C++
Prog: star.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// print a signed integer
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int inf;
struct _init_{
_init_(){
memset(&inf,0x3f,sizeof(inf));
}
};
_init_ ___INIT__;
set<pair<pii,int> > st;
int n,m,q;
int bit[100005],a[100005],s[100005],tot;
void add(int x,int y){
for(;x;x^=x&-x)bit[x]=bit[x]+y;
}
int sum(int x){
int y=0;
for(;x<=n;x+=x&-x)y+=bit[x];
return y;
}
void split(int x){
if(!x)return;
auto it=st.upper_bound(make_pair(make_pair(x,inf),inf));
--it;
if(it->x.y>x){
auto w=*it;
st.erase(it);
st.insert(make_pair(make_pair(w.x.x,x),w.y));
st.insert(make_pair(make_pair(x+1,w.x.y),w.y));
}
}
void rebuild(int l,int r,int w){
split(l-1);
split(r);
while(1){
auto it=st.lower_bound(make_pair(make_pair(l,0),0));
if(it==st.end())break;
if(it->x.x>r)break;
add(it->y,s[it->x.x-1]-s[it->x.y]);
st.erase(it);
}
add(w,s[r]-s[l-1]);
st.insert(make_pair(make_pair(l,r),w));
}
int qs[100005],qt[100005],ans[100005];
vector<pii> qu[100005];
vector<int> to[100005];
int siz[100005],hvy[100005],top[100005],dis[100005];
int dfn[100005],fa[100005];
int t;
void dfs(int x,int f){
siz[x]=1;
for(auto i:to[x])if(i!=f){
dfs(i,x);
siz[x]+=siz[i];
if(siz[hvy[x]]<siz[i])hvy[x]=i;
}
}
void dfs2(int x,int f){
dfn[x]=++t;
s[t]=s[t-1]+a[x];
if(hvy[x]){
top[t+1]=top[t];
dis[t+1]=dis[dfn[x]];
dfs2(hvy[x],x);
}
for(auto i:to[x])if(i!=f && i!=hvy[x]){
top[t+1]=t+1;
fa[t+1]=dfn[x];
dis[t+1]=dis[dfn[x]]+1;
dfs2(i,x);
}
}
void work(int u,int v,int x){
u=dfn[u];v=dfn[v];
while(1){
if(top[u]==top[v]){
if(u>v)swap(u,v);
rebuild(u,v,x);
return;
}
if(dis[u]<dis[v])swap(u,v);
rebuild(top[u],u,x);
u=fa[top[u]];
}
}
signed main(){
freopen("star.in","r",stdin);
freopen("star.out","w",stdout);
read(n,m,q);
for(int i=1;i<=n;++i){read(a[i]);tot+=a[i];}
st.insert(make_pair(make_pair(1,n),0));
int u,v;
for(int i=1;i<n;++i){
read(u,v);
to[u].push_back(v);
to[v].push_back(u);
}
dfs(1,0);
top[1]=1;
dfs2(1,0);
for(int i=1;i<=m;++i){
read(qs[i],qt[i]);
}
int l,r;
for(int i=1;i<=q;++i){
read(l,r);
qu[r].push_back(make_pair(l,i));
}
for(int i=1;i<=m;++i){
work(qs[i],qt[i],i);
for(auto j:qu[i])
ans[j.y]=sum(j.x);
}
for(int i=1;i<=q;++i)write(ans[i],'\n');
return 0;
}