题目链接
题意：
给一棵树。有边权
1、询问路径的边权和
2、询问沿着路径的第k个点标。
思路：lct裸题。
#include <iostream>
#include <fstream>
#include <string>
#include <time.h>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <stack>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if (x>9) pt(x / 10);
putchar(x % 10 + '0');
}
typedef long long ll;
typedef pair<int, int> pii;
const int N = 30005;
const int inf = 10000000;
struct Node *null;
struct Node{
Node *fa, *ch[2];
int size;
int val, ma, sum, id;
bool rev;
inline void put(){
printf("%d:id, %d,%d,%d (%d,%d) fa:%d \n", id, val, ma, sum, ch[0]->id, ch[1]->id, fa->id);
}
void debug(Node *x){
if (x == null)return;
x->put();
if (x->ch[0] != null)putchar('L'), debug(x->ch[0]);
if (x->ch[1] != null)putchar('r'), debug(x->ch[1]);
}
inline void clear(int _val, int _id){
fa = ch[0] = ch[1] = null;
size = 1;
rev = 0;
id = _id;
val = ma = sum = _val;
}
inline void add_val(int _val){
val += _val;
sum += _val;
ma = max(ma, val);
}
inline void push_up(){
size = 1 + ch[0]->size + ch[1]->size;
sum = ma = val;
if (ch[0] != null) {
sum += ch[0]->sum;
ma = max(ma, ch[0]->ma);
}
if (ch[1] != null){
sum += ch[1]->sum;
ma = max(ma, ch[1]->ma);
}
}
inline void push_down(){
if (rev){
flip(); ch[0]->rev ^= 1; ch[1]->rev ^= 1;
}
}
inline void setc(Node *p, int d){
ch[d] = p;
p->fa = this;
}
inline bool d(){
return fa->ch[1] == this;
}
inline bool isroot(){
return fa == null || fa->ch[0] != this && fa->ch[1] != this;
}
inline void flip(){
if (this == null)return;
swap(ch[0], ch[1]);
rev ^= 1;
}
inline void go(){//从链头開始更新到this
if (!isroot())fa->go();
push_down();
}
inline void rot(){
Node *f = fa, *ff = fa->fa;
int c = d(), cc = fa->d();
f->setc(ch[!c], c);
this->setc(f, !c);
if (ff->ch[cc] == f)ff->setc(this, cc);
else this->fa = ff;
f->push_up();
}
inline Node*splay(){
go();
while (!isroot()){
if (!fa->isroot())
d() == fa->d() ?
fa->rot() : rot();
rot();
}
push_up();
return this;
}
inline Node* access(){//access后this就是到根的一条splay，而且this已经是这个splay的根了
for (Node *p = this, *q = null; p != null; q = p, p = p->fa){
p->splay()->setc(q, 1);
p->push_up();
}
return splay();
}
inline Node* find_root(){
Node *x;
for (x = access(); x->push_down(), x->ch[0] != null; x = x->ch[0]);
return x;
}
void make_root(){
access()->flip();
}
void cut(){//把这个点的子树脱离出去
access();
ch[0]->fa = null;
ch[0] = null;
push_up();
}
void cut(Node *x){
if (this == x || find_root() != x->find_root())return;
else {
x->make_root();
cut();
}
}
void link(Node *x){
if (find_root() == x->find_root())return;
else {
make_root(); fa = x;
}
}
};
Node pool[N], *tail;
Node *node[N], *ee[N];
int n, q;
void debug(Node *x){
if (x == null)return;
x->put();
debug(x->ch[0]);
debug(x->ch[1]);
}
inline int ask(Node *x, Node *y){
x->access();
// for (int i = 1; i <= n; i++)debug(node[i]), putchar('\n');
for (x = null; y != null; x = y, y = y->fa){
y->splay();
// for (int i = 1; i <= n; i++)debug(node[i]), putchar('\n');
if (y->fa == null)return y->ch[1]->sum + x->sum;
y->setc(x, 1);
y->push_up();
}
}
inline Node* get_kth(Node *x, int k){
while (x->ch[0]->size + 1 != k){
if (x->ch[0]->size >= k)
x = x->ch[0];
else k -= x->ch[0]->size + 1, x = x->ch[1];
}
return x;
}
inline int query_kth(Node *x, Node *y, int k){
x->access();
for (x = null; y != null; x = y, y = y->fa){
y->splay();
if (y->fa == null){
if (k == y->ch[1]->size + 1)return y->id;
if (k < y->ch[1]->size + 1)return get_kth(y->ch[1], y->ch[1]->size - k + 1)->id;
return get_kth(x, k - y->ch[1]->size - 1)->id;
}
y->setc(x, 1);
y->push_up();
}
}
struct Edge{
int from, to, dis, id, nex;
}edge[N << 1];
int head[N], edgenum;
void add(int u, int v, int dis, int id){
Edge E = { u, v, dis, id, head[u] };
edge[edgenum] = E;
head[u] = edgenum++;
}
bool vis[N];
void bfs(){
fill(vis + 1, vis + 1 + n, false);
queue<int>q;
q.push(1);
vis[1] = true;
while (!q.empty()){
int u = q.front(); q.pop();
for (int i = head[u]; ~i; i = edge[i].nex){
int v = edge[i].to;
if (vis[v])continue;
vis[v] = true;
q.push(v);
ee[edge[i].id] = node[v];
node[v]->val = edge[i].dis;
node[v]->push_up();
node[v]->fa = node[u];
}
}
}
int main(){
int T; rd(T);
while (T--){
rd(n);
fill(head + 1, head + n + 1, -1); edgenum = 0;
tail = pool;
null = tail++;
null->clear(-inf, 0);
null->size = 0; null->sum = 0;
for (int i = 1; i <= n; i++) {
node[i] = tail++;
node[i]->clear(0, i);
}
for (int i = 1, u, v, d; i < n; i++){
rd(u); rd(v); rd(d);
add(u, v, d, i);
add(v, u, d, i);
}
bfs();
char str[10]; int u, v, k;
while (true){
scanf("%s", str);
if (str[1] == 'O')break;
rd(u); rd(v);
if (str[0] == 'D')pt(ask(node[u], node[v])), putchar('\n');
else {
rd(k);
pt(query_kth(node[u], node[v], k)); putchar('\n');
}
}
puts("");
}
return 0;
}
/*
1
6
1 2 1
2 4 1
2 5 2
1 3 1
3 6 2
DIST 4 6
KTH 4 6 4
KTH 6 5 4
DIST 2 5
*/