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游戏文明策略模拟

最新推荐文章于 2025-01-20 09:00:00 发布

原创最新推荐文章于 2025-01-20 09:00:00 发布 · 923 阅读

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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 citiesai 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.

Sample test(s)

input
6 0 6
2 1 2
2 3 4
2 5 6
2 3 2
2 5 3
1 1

output
4


用半个小时A完了abc，一个半小时写这题都没写出来，真是拙计。。先是想找重心，然后维护每棵子树的半径，直径，和重心的位置。不会树的重心呀。。。套模板改来改去比赛就这么tm结束了。。。赛后有人说只用维护树的直径，仔细一想，是的呀！我真是sb了。注意：两棵子树合并后的树的直径，不能只考虑用两棵子树的半径相加，有可能合并后还是原来某棵子树的直径。
#include<cstdio>
#include<map>
#include<queue>
#include<cstring>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
const int maxn = 300000 + 5;
const int INF = 1000000000;
typedef long long LL;
typedef pair<int, int> P;
#define fi first
#define se second

vector<int> G[maxn];
int n;
int fa[maxn], diameter[maxn], vis[maxn];
int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}

struct Node{
    int x, from, cnt;
    Node(int x, int from, int cnt){
        this -> x = x;
        this -> from = from;
        this -> cnt = cnt;
    }
};
queue<Node> q;

P bfs(int x){
    while(!q.empty())
        q.pop();
    q.push(Node(x, -1, 0));
    int pos, from, cnt;
    while(!q.empty()){
        Node tem = q.front();
        q.pop();
        pos = tem.x;
        from = tem.from;
        cnt = tem.cnt;
        vis[pos] = 1;
        for(int i = 0;i < G[pos].size();i++){
            int to = G[pos][i];
            if(to != from){
                q.push(Node(to, pos, cnt+1));
            }
        }
    }

    return P(cnt, pos);
}

void pre(){
    memset(vis, 0, sizeof vis);
    for(int i = 1;i <= n;i++){
        if(vis[i] == 0){
            int tem = bfs(i).se;
            int X = Find(i);
            diameter[X] = bfs(tem).fi;
        }
    }
}

int query(int x){
    return diameter[Find(x)];
}

void join(int x, int y){
    int X = Find(x);
    int Y = Find(y);
    if(X != Y){
        int radx = (diameter[X]+1)/2;
        int rady = (diameter[Y]+1)/2;
        fa[X] = Y;
        diameter[Y] = max(radx+rady+1, max(diameter[X],diameter[Y]));
    }
}

int main(){
    int m, q;
    while(scanf("%d%d%d", &n, &m, &q) != EOF){
        for(int i = 0;i < maxn;i++){
            G[i].clear();
            fa[i] = i;
        }
        while(m--){
            int x, y;
            scanf("%d%d", &x, &y);
            G[x].push_back(y);
            G[y].push_back(x);
            int X = Find(x);
            int Y = Find(y);
            fa[X] = Y;
        }

        pre();

        while(q--){
            int kind;
            scanf("%d", &kind);
            if(kind == 1){
                int x;
                scanf("%d", &x);
                cout << query(x) << endl;
            }
            else{
                int x, y;
                scanf("%d%d", &x, &y);
                join(x, y);
            }
        }
    }
    return 0;
}