题目链接：
题目
C. Civilization
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
问题描述
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.
输入
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.
输出
For each event of the first type print the answer on a separate line.
样例
input
6 0 6
2 1 2
2 3 4
2 5 6
2 3 2
2 5 3
1 1
output
4
题意
给你一个森林，进行两个操作：
- 输入一个x，求x所在的树的直径
- 输入u，v，把u，v所在的树合并起来
题解
求出每颗树的中心，作为集合的中心。
两颗树合并，只要把直径比较短的树的中心合并到直径比较长的树的中心就可以了。
合并完之后的直径长度为max(r1,r2,(r1+1)/2+(r2+1)/2+1)。其中r1为第一颗树的直径，r2为第二颗树的直径
没写过树的中心，写搓了：
先求树的直径，然后找直径的中间点。连续dfs了三次。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 3e5+10;
int n, m, q;
int fa[maxn],r[maxn];
int find(int x) {
return fa[x] = fa[x] == x ? x : find(fa[x]);
}
vector<int> G[maxn];
int vis[maxn];
void dfs(int u, int p, int &len, int &x,int d) {
if (len < d) x = u, len = d;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v == p) continue;
dfs(v, u, len, x, d + 1);
}
}
bool dfs2(int u, int p, int y, int len,int dep,int &ctr) {
bool ok = 0;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v == p) continue;
ok=dfs2(v, u, y, len,dep+1, ctr);
if (ok&&dep == len / 2) ctr = u;
if (ok) return ok;
}
if (u == y) ok = 1;
return ok;
}
void dfs3(int u, int p,int ctr) {
vis[u] = 1;
fa[u] = ctr;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v == p) continue;
dfs3(v, u, ctr);
}
}
void init() {
for (int i = 0; i <= n; i++) fa[i] = i,r[i]=0;
memset(vis, 0, sizeof(vis));
}
int main() {
scanf("%d%d%d", &n, &m, &q);
init();
int cmd, u,v,x;
while (m--) {
scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
int ctr, l, x, y;
l = -1; dfs(i, -1, l, x, 0);
l = -1; dfs(x, -1, l, y, 0);
ctr = i;
dfs2(x, -1, y, l,0,ctr);
r[ctr] = l;
dfs3(i, -1,ctr);
}
}
while (q--) {
scanf("%d", &cmd);
if (cmd == 1) {
scanf("%d", &x);
int px = find(x);
printf("%d
", r[px]);
}
else {
scanf("%d%d", &u, &v);
int pu = find(u), pv = find(v);
if (pu != pv) {
int maxr = max(r[pu], r[pv]);
maxr = max(maxr, (r[pu] + 1) / 2 + (r[pv] + 1) / 2 + 1);
if (r[pu] > r[pv]) fa[pv] = pu, r[pu] = maxr;
else fa[pu] = pv, r[pv] = maxr;
}
}
}
return 0;
}
发现只要求直径就可以了，随便找个点为根，不用找树中心。。orz。。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 3e5+10;
int n, m, q;
int fa[maxn],r[maxn];
int find(int x) {
return fa[x] = fa[x] == x ? x : find(fa[x]);
}
vector<int> G[maxn];
int rt;
void dfs(int u, int p, int &len, int &x,int d) {
fa[u] = rt;
if (len < d) x = u, len = d;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v == p) continue;
dfs(v, u, len, x, d + 1);
}
}
void init() {
for (int i = 0; i <= n; i++) fa[i] = i,r[i]=0;
}
int main() {
scanf("%d%d%d", &n, &m, &q);
init();
int cmd, u,v,x;
while (m--) {
scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
for (int i = 1; i <= n; i++) {
if (fa[i]==i) {
int l, x, y;
rt = i;
l = -1; dfs(i, -1, l, x, 0);
l = -1; dfs(x, -1, l, y, 0);
r[i] = l;
}
}
while (q--) {
scanf("%d", &cmd);
if (cmd == 1) {
scanf("%d", &x);
int px = find(x);
printf("%d
", r[px]);
}
else {
scanf("%d%d", &u, &v);
int pu = find(u), pv = find(v);
if (pu != pv) {
int maxr = max(r[pu], r[pv]);
maxr = max(maxr, (r[pu] + 1) / 2 + (r[pv] + 1) / 2 + 1);
if (r[pu] > r[pv]) fa[pv] = pu, r[pu] = maxr;
else fa[pu] = pv, r[pv] = maxr;
}
}
}
return 0;
}