time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Andrew plays a game called “Civilization”. Dima helps him.

The game has n cities and m bidirectional roads. The cities are numbered from 1 to n. Between any pair of cities there either is a single (unique) path, or there is no path at all. A path is such a sequence of distinct cities v1, v2, …, vk, that there is a road between any contiguous cities vi and vi + 1 (1 ≤ i < k). The length of the described path equals to (k - 1). We assume that two cities lie in the same region if and only if, there is a path connecting these two cities.

During the game events of two types take place:

Andrew asks Dima about the length of the longest path in the region where city x lies.
Andrew asks Dima to merge the region where city x lies with the region where city y lies. If the cities lie in the same region, then no merging is needed. Otherwise, you need to merge the regions as follows: choose a city from the first region, a city from the second region and connect them by a road so as to minimize the length of the longest path in the resulting region. If there are multiple ways to do so, you are allowed to choose any of them.
Dima finds it hard to execute Andrew’s queries, so he asks you to help him. Help Dima.

Input
The first line contains three integers n, m, q (1 ≤ n ≤ 3·105; 0 ≤ m < n; 1 ≤ q ≤ 3·105) — the number of cities, the number of the roads we already have and the number of queries, correspondingly.

Each of the following m lines contains two integers, ai and bi (ai ≠ bi; 1 ≤ ai, bi ≤ n). These numbers represent the road between cities ai and bi. There can be at most one road between two cities.

Each of the following q lines contains one of the two events in the following format:

1 xi. It is the request Andrew gives to Dima to find the length of the maximum path in the region that contains city xi (1 ≤ xi ≤ n).
2 xi yi. It is the request Andrew gives to Dima to merge the region that contains city xi and the region that contains city yi (1 ≤ xi, yi ≤ n). Note, that xi can be equal to yi.
Output
For each event of the first type print the answer on a separate line.

Examples
inputCopy
6 0 6
2 1 2
2 3 4
2 5 6
2 3 2
2 5 3
1 1
outputCopy
4

题目大意：给定N，M和Q，N表示有N个城市，M条已经修好的路，修好的路是不能改变的，然后是Q次操作，操作分为两种，一种是查询城市x所在的联通集合中，最长的路为多长。二是连接两个联通集合，采用联通之后最长路最短的方案。
解析:
对于建好的图
进行两次dfs,第一次dfs得到最大边的一个端点,
然后以这个端点为根节点,进行第二次dfs,将这个连通块中的所有点的pre设为这个点.并更新这个连通块最长路的值
Q次操作，用并查集维护(在这里请注意并查集的写法,比最基本的写法要效率要高)，联通的时候要采用最长路径最短的方案，所以s的转移方程变为s = max(s, (s+1)/2 + (s0+1)/2 + 1),这个公式很好理解就不多讲了.

#include<bits/stdc++.h>
using namespace  std;
#define ll long long
#define pb push_back
#define inf 2099999999
#define mod 1000000007
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep1(i,a,b) for(int i=a;i>=b;i--)
const int N=3e5+100;
int arr[N];
int pre[N];
int dis[N];
vector<int>G[N];
int ans,rec;
int fi(int x)
{
    return x==pre[x]?x:pre[x]=fi(pre[x]);
}
void dfs(int now ,int last,int root,int d)
{
    pre[now]=root;
    if(d>ans) rec=now,ans=d;
    for(int i=0;i<G[now].size();i++)
    {
        int y =G[now][i];
        if(y!=last)
            dfs(y,now ,root,d+1);
    }
}
int main()
{
    #ifdef LOCAL_DEFINE
        freopen("D://rush.txt", "r", stdin);
    #endif
    //ios::sync_with_stdio(false),cin.tie(0);
    int n,m,q,a,b,x,y;
    scanf("%d%d%d",&n,&m,&q);
    rep(i,1,n) pre[i]=i;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&a,&b);
        G[a].pb(b);
        G[b].pb(a);
    }
    for(int i=1;i<=n;i++)
    {
        if(pre[i]==i)
        {
            ans=0;
            dfs(i,0,i,0);
            dfs(rec,0,rec,0);
            dis[rec]=ans;
        }
    }
    while(q--)
    {
        scanf("%d",&a);
        if(a==1)
        {
            scanf("%d",&x);
            printf("%d\n", dis[fi(x)]);
        }
        else
        {
            scanf("%d%d",&x,&y);
            x=fi(x),y=fi(y);
            if(x!=y)
            {
                pre[y]=x;
                if(dis[x]<dis[y]) swap(dis[x],dis[y]);
                dis[x]=max(dis[x],(dis[x]+1)/2+(dis[y]+1)/2+1);
            }
        }

    }
    return 0;
}