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 1000000int 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;}
