学步园

题目：Codeforces Round #260 (Div. 2) E. Civilization

题意：输入n,m,q，n代表点数，m是边数，q是问题数。问题有两种一种是链接两个集合使得合并后的最长路径最短，另一种是求某个集合的最长路径长度。

思路：据说这个叫树的直径，求法就是两次dfs。要合并后的直径最短，合并点一定是两个集合最长链的中点。

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <string>
using namespace std;
#define For(i,a) for(i=0;i<a;i++)
#define Foru(i,a,b) for(i=a;i<=b;i++)
#define Ford(i,a,b) for(i=a;i>=b;i--)
#define clr(ar,vel) memset(ar,vel,sizeof(ar))
#define PB push_back
#define maxint 0x7fffffff

const int maxn = 300010;
vector <int> G[maxn];
int diameter[maxn];
int parent[maxn];
int vis[maxn];
int NODE, dep;
int getSet(int x){
	return parent[x] = x == parent[x] ? x : getSet(parent[x]);
}
int mergSet(int x, int y){
	x = getSet(x);
	y = getSet(y);
	if( x > y ) swap(x,y);
	parent[y] = x;
}
int dfs(int x, int d, int p){
//	cout << x << ' ' << d << endl;
	if( d > dep) {
		dep = d;
		NODE = x;
	}
	for(int i = 0; i < G[x].size(); i ++){
//		cout << "ok" << endl;
//		system("pause");
		if( G[x][i] != p) dfs(G[x][i], d+1, x);
	}
}
void solve(int n){
	for(int i = 1; i <= n; i ++) if(!vis[getSet(i)]){
		dep = -1;
		dfs(i, 0, -1);
		dfs(NODE, 0, -1);
		
//		cout << dep << endl;
		vis[getSet(i)] = 1;
		diameter[getSet(i)] = dep;
	}
}
void init(int n){
	for(int i = 0; i <= n; i ++) {
		parent[i] = i;
		diameter[i] = 0;
		G[i].clear();
		vis[i] = 0;
	}
}
int main(){
	int n, m, q;
	int from, to, x;
	scanf("%d%d%d",&n,&m,&q);
	init(n);
	for(int i = 0; i < m; i ++) {
		scanf("%d%d",&from,&to);
		G[from].push_back(to);
		G[to].push_back(from);
		mergSet(from, to);
	}
	solve(n);
	while(q--){
		scanf("%d",&x);
		if( x == 1) {
			scanf("%d",&from);
			printf("%d\n",diameter[getSet(from)]);
		}
		else {
			scanf("%d%d",&from,&to);
			from = getSet(from);
			to = getSet(to);
			if( from == to ) continue;
			int newdia = max(diameter[from], diameter[to]);
			newdia = max(newdia, (diameter[from]+1)/2+(diameter[to]+1)/2+1);
			mergSet(from, to);
			diameter[getSet(from)] = newdia;
		}
	}
	return 0;
}