题解
点分治,然后每个点上挂着一个距离不超过(a_{i})的颜色改成(c)
用一个单调栈维护距离单调递减,每次查询在每个包括这个点的分治中心的单调栈上二分,找到修改最靠前的颜色作为这个点的颜色
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct ch {
int id,d,c;
};
struct node {
int to,next,val;
}E[MAXN * 2];
int N,head[MAXN],sumE,d[MAXN],M;
bool vis[MAXN];
vector<int> aux[MAXN],dep[MAXN],poi;
vector<ch> sta[MAXN];
void add(int u,int v,int c) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].val = c;
head[u] = sumE;
}
int Calc_G(int st) {
static int fa[MAXN],son[MAXN],siz[MAXN],que[MAXN],ql,qr;
ql = 1,qr = 0;
que[++qr] = st;son[st] = 0;siz[st] = 1;fa[st] = 0;
while(ql <= qr) {
int u = que[ql++];
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v] && v != fa[u]) {
que[++qr] = v;
siz[v] = 1;son[v] = 0;fa[v] = u;
}
}
}
int res = que[qr];
for(int i = qr ; i >= 1 ; --i) {
int u = que[i];
if(fa[u]) {
siz[fa[u]] += siz[u];
son[fa[u]] = max(son[fa[u]],siz[u]);
}
son[u] = max(son[u],qr - siz[u]);
if(son[u] < son[res]) res = u;
}
return res;
}
void dfs_for_dep(int u,int fa) {
poi.pb(u);
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v] && v != fa) {
d[v] = d[u] + E[i].val;
dfs_for_dep(v,u);
}
}
}
void dfs_divide(int u) {
int G = Calc_G(u);
vis[G] = 1;
sta[G].pb((ch){0,1000000000,0});
d[G] = 0;poi.clear();
dfs_for_dep(G,0);
for(int i = 0 ; i < poi.size() ; ++i) {
aux[poi[i]].pb(G);
dep[poi[i]].pb(d[poi[i]]);
}
for(int i = head[G] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) dfs_divide(v);
}
}
void Init() {
read(N);
int u,v,c;
for(int i = 1 ; i < N ; ++i) {
read(u);read(v);read(c);
add(u,v,c);add(v,u,c);
}
dfs_divide(1);
}
void Change(int id,int x,int d,int c) {
for(int i = aux[x].size() - 1 ; i >= 0 ; --i) {
int u = aux[x][i],t = d - dep[x][i];
if(t < 0) continue;
while(sta[u].size()) {
ch l = sta[u].back();
if(l.d <= t) sta[u].pop_back();
else break;
}
sta[u].pb((ch){id,t,c});
}
}
int Query(int x) {
int id = 0,c = 0;
for(int i = aux[x].size() - 1 ; i >= 0 ; --i) {
int u = aux[x][i],d = dep[x][i];
int L = 0,R = sta[u].size() - 1;
while(L < R) {
int mid = (L + R + 1) >> 1;
if(sta[u][mid].d >= d) L = mid;
else R = mid - 1;
}
if(sta[u][L].id > id) {id = sta[u][L].id;c = sta[u][L].c;}
}
return c;
}
void Solve() {
read(M);
int op,v,d,c;
for(int i = 1 ; i <= M ; ++i) {
read(op);read(v);
if(op == 1) {
read(d);read(c);
Change(i,v,d,c);
}
else {
out(Query(v));enter;
}
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}