455C - Civilization

Solution:

An nice problem, but my implementation is a bit messy :P

The most important observation that greatly simplify the problem:

If node u has the property such that for every v reachable from u, \(max_{v \text{ reachable from } u} \{ d(u,v) \} \) is minimum, then u must be on the longest path in that connected component. Furthermore, u must be on the middle of that path.

Firstly, why do we want to find such node u? We can think a node with that kind of property as the "midpoint" or "center" of the whole tree, where d(u,v) will be analogous to the "radius" of the tree. So when we have to merge to connected component, all we need to do is to connect the two "centers" of the connected components, and it is guaranteed that the resulting connected component will have the minimum longest path possible.

Next, the fact that u will always be on the mid point of the longest path means to find the length of longest path after merging two different components can be done in O(1) by adding up the "radius" of both components, and similarly the "radius" of the resulting connected component can be computed by just halving the length of its longest path.

To support merging efficiently, we can use disjoint set union structure to represent each connected component, since the information about the underlying graph itself is not important, as only the "radius" and the connectedness that matter.

#include <iostream> #include <cstdio> #include <algorithm> #include <vector> using namespace std; /* Lemma: if u is the node such that max{d(u,v)} is minimum, then u is on the longest path And u must be the one on the middle. */ int N,M,Q; vector<vector<int> > adj; int par[300005]; int val[300005][2]; int dp[300005]; int len[300005]; int vis[300005]; int rad; int get_par(int i){ return (par[i] == i ? i : (par[i] = get_par(par[i]))); } void merge_set(int u, int v){ int x = get_par(u); int y = get_par(v); if(x==y)return; int r = max(val[x][0],val[x][1]) + max(val[y][0],val[y][1]) + 1; r = max(r, val[x][0]+val[x][1]); r = max(r, val[y][0]+val[y][1]); val[y][0] = r/2; val[y][1] = r - r/2; par[x] = y; } void dfs(int u, int depth){ vis[u] = 1; len[u] = 0; vector<int> tmp; for(int i=0;i<adj[u].size();++i){ int v = adj[u][i]; if(vis[v])continue; dfs(v,depth+1); len[u] = max(len[u], len[v]+1); tmp.push_back(len[v]); rad = max(rad, depth+1+len[v]); } sort(tmp.begin(),tmp.end()); if(tmp.size()>1){ rad = max(rad, 2 + tmp[tmp.size()-1] + tmp[tmp.size()-2]); } } int main(){ int u,v,w; scanf("%d%d%d", &N,&M,&Q); adj = vector<vector<int> > (N+3); for(int i=1;i<=N;++i)par[i]=i; for(int i=0;i<M;++i){ scanf("%d%d",&u,&v); merge_set(u,v); adj[u].push_back(v); adj[v].push_back(u); } for(int i=1;i<=N;++i){ if(vis[i])continue; rad = 0; dfs(i, 0); v = get_par(i); val[v][0] = rad/2; val[v][1] = rad - rad/2; } for(int qq=0;qq<Q;++qq){ scanf("%d", &w); if(w==1){ scanf("%d",&u); int x = get_par(u); printf("%d\n",val[x][0]+val[x][1]); } else { scanf("%d%d",&u,&v); merge_set(u,v); } } return 0; }