Codeforces 455c civilization tree diameter + query set

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Question:

K queries for graphs with n vertices and M undirected Edges
No duplicate edge, auto ring, or ring
Define two points to belong to one country: when these two points are connected
Operation 1 X: output the longest route length in the country where X is located

Operation 2 x Y: Ignore if x y belongs to a country
If it does not belong to a country, an edge is connected between the two sets to minimize the longest link after the connection.

The longest path of two sets. The first is to find the midpoint of the longest path of the Two Sets for connection.

Then the longest path length after connection is path [x]/2 + path [y]/2 + 1 (except for 2 rounded up)

Then we can prepare the diameter of each tree.

Because you only need to find the length of the tree's diameter and do not care about the path of the tree's diameter

Therefore, for all the sub-trees of BFS (x), the length of the two longest sub-trees is the diameter. BFS: once

#include <cstdio>#include <cstring>#include<iostream>#include <queue>#include <set>using namespace std;#define inf 10000000#define N 300005struct Edge{int to, nex;}edge[N<<1];int head[N], edgenum;void add(int u, int v){Edge E = {v, head[u]};edge[edgenum] = E;head[u] = edgenum++;}int f[N], path[N];int find(int x){return x==f[x]?x:f[x] = find(f[x]);}void Union(int x, int y){int fx = find(x), fy = find(y);if(fx == fy)return ;if(fx>fy)swap(fx, fy);f[fx] = fy;int now = path[fx]/2 + path[fy]/2 +1;if(path[fx]&1)now++;if(path[fy]&1)now++;path[fx] = path[fy] = max(max(path[fx], path[fy]), now);}int n, m;int dis[N];vector<int>G[N];int BFS(int x){ int E = x;queue<int>q;  for(int i = 0; i < G[f[x]].size(); i++)dis[G[f[x]][i]] = inf;q.push(x);dis[x]=0;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(dis[v] > dis[u]+1){dis[v] = dis[u]+1;if(dis[v]>dis[E])E = v;q.push(v);}}}return E;  }void work(int x){int S = BFS(x);S = BFS(S);path[x] = dis[S];}set<int>s;void init(){s.clear();memset(head, -1, sizeof head); edgenum = 0;memset(path, 0, sizeof path);for(int i = 0; i <= n; i++)f[i] = i, G[i].clear();}int main(){  int i, u, v, q, op;while(cin>>n>>m>>q){init();while(m--){scanf("%d %d",&u,&v);add(u,v);add(v,u);Union(u,v);}for(i = 1; i <= n; i++)find(i);for(i = 1; i <= n; i++) {s.insert(f[i]);G[f[i]].push_back(i);}for(set<int>::iterator it = s.begin(); it!=s.end(); it++)work(*it);while(q--){scanf("%d %d",&op, &u);if(op==1){u = find(u);printf("%d\n", path[u]);}else {scanf("%d",&v);Union(u, v);}}}return 0;}