[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=1095
[算法]
首先建出点分树，然后每一个点开两个堆。“第一个堆记录子树中所有节点到父亲节点的距离 ，第二个堆记录所有子节点的堆顶 ，那么一个节点的堆2中的最大和次大加起来就是子树中经过这个节点的最长链。然后我们最后开一个全局的堆，记录所有堆2中最大值和次大值之和。那么全局的堆顶就是答
案
时间复杂度 ： O(NlogN ^ 2)
[代码]
#include<bits/stdc++.h> using namespace std; #define N 100010 #define M 500010 typedef long long ll; typedef long double ld; typedef unsigned long long ull; int n , root , tot , timer; int size[N] , head[N] , weight[N] , father[N] , ans[N]; bool visited[N] , light[N]; map< int , pair<int , int> > mp[N]; struct edge { int to , nxt; } e[M << 1]; struct Heap { priority_queue< int > hp , era; inline void ins(int value) { hp.push(value); } inline void del(int value) { if (value != 0) era.push(value); } inline int max() { while (!hp.empty() && !era.empty() && era.top() == hp.top()) { hp.pop(); era.pop(); } if (hp.empty()) return 0; else return hp.top(); } inline int secmax() { int tmp = max(); del(tmp); int ret = max(); ins(tmp); return ret; } } A , B[N] , C[N << 1]; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0'; x *= f; } inline void addedge(int u , int v) { ++tot; e[tot] = (edge){v , head[u]}; head[u] = tot; } inline void getroot(int u , int father , int total) { size[u] = 1; weight[u] = 0; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v == father || visited[v]) continue; getroot(v , u , total); size[u] += size[v]; chkmax(weight[u] , size[v]); } chkmax(weight[u] , total - size[u]); if (weight[u] < weight[root]) root = u; } inline void dfs(int u , int father , int rt , int belong , int depth) { mp[rt][u] = make_pair(belong , depth); C[belong].ins(depth); for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (v == father || visited[v]) continue; dfs(v , u , rt , belong , depth + 1); } } inline void work(int u) { visited[u] = true; for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (!visited[v]) { dfs(v , u , u , ++timer , 1); B[u].ins(C[timer].max()); } } ans[u] = B[u].max() + B[u].secmax(); A.ins(ans[u]); for (int i = head[u]; i; i = e[i].nxt) { int v = e[i].to; if (!visited[v]) { root = 0; getroot(v , u , size[v]); father[root] = u; work(root); } } } int main() { read(n); for (int i = 1; i < n; i++) { int x , y; read(x); read(y); addedge(x , y); addedge(y , x); } int q; read(q); root = 0; weight[0] = n; getroot(1 , 0 , n); work(root); int cnt = n; while (q--) { char type[5]; scanf("%s" , type); if (type[0] == 'C') { int x , y; scanf("%d" , &x); y = x; if (light[x]) { light[x] = false; ++cnt; A.del(ans[x]); ans[x] = B[x].max() + B[x].secmax(); A.ins(ans[x]); x = father[x]; while (x != 0) { pair<int , int> tmp = mp[x][y]; B[x].del(C[tmp.first].max()); C[tmp.first].ins(tmp.second); B[x].ins(C[tmp.first].max()); A.del(ans[x]); if (light[x]) { if (B[x].secmax()) ans[x] = B[x].max() + B[x].secmax(); else ans[x] = 0; } else ans[x] = B[x].max() + B[x].secmax(); A.ins(ans[x]); x = father[x]; } } else { light[x] = true; --cnt; A.del(ans[x]); if (B[x].secmax()) ans[x] = B[x].max() + B[x].secmax(); else ans[x] = 0; A.ins(ans[x]); x = father[x]; while (x != 0) { pair<int , int> tmp = mp[x][y]; B[x].del(C[tmp.first].max()); C[tmp.first].del(tmp.second); B[x].ins(C[tmp.first].max()); A.del(ans[x]); if (light[x]) { if (B[x].secmax()) ans[x] = B[x].max() + B[x].secmax(); else ans[x] = 0; } else ans[x] = B[x].max() + B[x].secmax(); A.ins(ans[x]); x = father[x]; } } } else { if (!cnt) printf("-1 "); else if (cnt == 1) printf("0 "); else printf("%d " , A.max()); } } return 0; }