1、题目大意：就是在动态的树上路径权值第k大。
2、分析：这个就是树链剖分+树套树
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 1000000
int Height[M], Top[M], value[M], num[M], Size[M], Fa[M];
int ST_tot, tot;
int son[M], head[M], Next[M];
int n, m;
struct Node{
Node *ch[2];
int cnt, num, r, v;
bool operator < (const Node& rhs) const{
return r < rhs.r;
}
int cmp(int x){
if(x == v) return -1;
if(x < v) return 0;
return 1;
}
void maintain(){
cnt = num;
if(ch[0]) cnt += ch[0] -> cnt;
if(ch[1]) cnt += ch[1] -> cnt;
}
} *root[2 * M], ft[5 * M];
int treap_tot;
inline void treap_rotate(Node* &o, int d){
Node* k = o -> ch[d ^ 1];
o -> ch[d ^ 1] = k -> ch[d];
k -> ch[d] = o;
o -> maintain();
k -> maintain();
o = k;
return;
}
inline void treap_insert(Node* &o, int x){
if(o == NULL){
o = &ft[treap_tot ++];
o -> ch[0] = o -> ch[1] = NULL;
o -> cnt = o -> num = 1;
o -> v = x;
o -> r = rand();
}
else{
int d = o -> cmp(x);
if(d == -1){
o -> num ++;
}
else{
treap_insert(o -> ch[d], x);
if(o < o -> ch[d]) treap_rotate(o, d ^ 1);
}
}
o -> maintain();
}
inline void treap_remove(Node* &o, int x){
int d = o -> cmp(x);
if(d == -1){
if(o -> num > 1) o -> num --;
else if(o -> ch[0] == NULL) o = o -> ch[1];
else if(o -> ch[1] == NULL) o = o -> ch[0];
else {
int d2;
if(o -> ch[0] > o -> ch[1]) d2 = 1;
else d2 = 0;
treap_rotate(o, d2);
treap_remove(o -> ch[d2], x);
}
}
else treap_remove(o -> ch[d], x);
if(o) o -> maintain();
}
inline int treap_lessk(Node* &o, int k){
if(o == NULL) return 0;
int d = o -> cmp(k);
if(d == -1){
int ret = 0;
if(o -> ch[0]) ret += o -> ch[0] -> cnt;
return ret;
}
else if(d == 0){
return treap_lessk(o -> ch[0], k);
}
else{
int ss = o -> num;
if(o -> ch[0]) ss += o -> ch[0] -> cnt;
return treap_lessk(o -> ch[1], k) + ss;
}
}
inline void init(){
Top[1] = 1;
memset(head, -1, sizeof(head));
tot = ST_tot = 0;
}
inline void add(int l, int r, int o, int x, int y, int z){
if(y != -1) treap_remove(root[o], y);
treap_insert(root[o], z);
if(l == r) {
return;
}
int mid = (l + r) / 2;
if(x <= mid) add(l, mid, 2 * o, x, y, z);
else add(mid + 1, r, 2 * o + 1, x, y, z);
}
inline int query(int l, int r, int o, int x, int y, int z){
if(x <= l && r <= y) return root[o] -> cnt - treap_lessk(root[o], z);
int ret = 0, mid = (l + r) / 2;
if(x <= mid) ret += query(l, mid, 2 * o, x, y, z);
if(y > mid) ret += query(mid + 1, r, 2 * o + 1, x, y, z);
return ret;
}
inline void insert(int x, int y){
tot ++;
son[tot] = y;
Next[tot] = head[x];
head[x] = tot;
}
inline void dfs1(int x, int fa, int height){
Fa[x] = fa;
Height[x] = height;
Size[x] = 1;
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa){
dfs1(son[i], x, height + 1);
Size[x] += Size[son[i]];
}
}
inline void dfs2(int x, int fa){
++ ST_tot;
num[x] = ST_tot;
add(1, n, 1, ST_tot, -1, value[x]);
int o = 0, ss = 0;
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa){
if(Size[son[i]] > ss){
ss = Size[son[i]];
o = i;
}
}
if(o != 0){
Top[son[o]] = Top[x];
dfs2(son[o], x);
}
for(int i = head[x]; i != -1; i = Next[i]) if(son[i] != fa && o != i){
Top[son[i]] = son[i];
dfs2(son[i], x);
}
}
inline void real_add(int x, int y){
add(1, n, 1, num[x], value[x], y);
value[x] = y;
}
inline int check(int x, int y, int k){
int ret = 0;
while(Top[x] != Top[y]){
if(Height[Top[x]] < Height[Top[y]]) swap(x, y);
ret += query(1, n, 1, num[Top[x]], num[x], k);
x = Fa[Top[x]];
}
if(Height[x] < Height[y]) swap(x, y);
ret += query(1, n, 1, num[y], num[x], k);
return ret;
}
inline int real_query(int x, int y, int k){
int l = -1, r = 100000000;
while(l < r){
int mid = (l + r) / 2;
if(mid == l) mid ++;
if(check(x, y, mid) >= k) l = mid;
else r = mid - 1;
}
if(l == -1) return -1;
return l;
}
int main(){
scanf("%d%d", &n, &m);
init();
for(int i = 1; i <= n; i ++) scanf("%d", &value[i]);
for(int i = 1; i < n; i ++){
int x, y;
scanf("%d%d", &x, &y);
insert(x, y);
insert(y, x);
}
dfs1(1, 0, 1);
dfs2(1, 0);
for(int i = 1; i <= m; i ++){
int k, x, y;
scanf("%d%d%d", &k, &x, &y);
if(k == 0){
real_add(x, y);
}
else if(k > 0){
int qq = real_query(x, y, k);
if(qq == -1){
printf("invalid request!\n");
}
else{
printf("%d\n", qq);
}
}
}
return 0;
}