CodeFroces gym 100781 A.Adjoin the Networks（贪心）

题意：给出一张图，存在多棵树，你要把这些树连起来，形成一棵树，令这棵树的直径最小。

解法：很容易想到贪心一下就行了，找出直径最长的树的根作为连起来后的树的根，其他树的根连到这个根即可。然后求一下树的直径。难点在于怎么规定一个树的根，使得这棵树的深度最小。下面写的是类似于LCA两个端点向上走的方法，有dalao告诉我可以树形dp+dfs找。这都是可以的。但是注意：树的重心并不一定是这个最优的根。

代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
#include<stack>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
using namespace std;
typedef pair <int, int> pii;
const int maxn = 1e5 + 5;
int head[maxn], to[maxn << 1], nx[maxn << 1], tot;
bool vst[maxn];
int Min = 0x3f3f3f3f, key, dis[maxn], pre[maxn];
int n, l, mycnt;
int path1[maxn], path2[maxn], path[maxn], cnt1 = 0, cnt2 = 0, cnt = 0;void add_edge(int u, int v) {to[tot] = v, nx[tot] = head[u], head[u] = tot++;swap(u, v);to[tot] = v, nx[tot] = head[u], head[u] = tot++;
}vector <pii> vec;
int Max_dis;void dfs(int u, int par, int _dis) {vst[u] = 1; pre[u] = par;mycnt++;dis[u] = _dis;for(int i = head[u]; ~i; i = nx[i]) {int v = to[i];if(v != par) {dfs(v, u, _dis + 1);}}if(_dis > Max_dis) {key = u;Max_dis = _dis;}
}int cal(int x, int y) {cnt1 = cnt2 = cnt = 0;while (dis[x] > dis[y]) {path1[cnt1++] = x; x = pre[x];}while (dis[y] > dis[x]) {path2[cnt2++] = y; y = pre[y];}while (x != y) {path1[cnt1++] = x; x = pre[x];path2[cnt2++] = y; y = pre[y];}for (int i = 0; i < cnt1; ++i) {  path[cnt++] = path1[i];}path[cnt++] = x;for (int i = cnt2 - 1; i >= 0; --i) {  path[cnt++] = path2[i];}
//  for (int i = 0; i < cnt; ++i) {
//      printf("%d ", path[i]);
//  }
//  printf("\n");return path[cnt / 2];
}int main() {
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
#endifmemset(head, -1, sizeof(head));tot = 0;scanf("%d%d", &n, &l);for(int i = 0, u, v; i < l; i++) {scanf("%d%d", &u, &v);add_edge(u, v);}for(int i = 0; i < n; i++) {if(!vst[i]) {Max_dis = -1; int t1, t2, t3;dfs(i, i, 0);Max_dis = -1;mycnt = 0;t1 = key; dfs(key, key, 0); t2 = key; t3 = cal(t1, t2);if(mycnt != 1)vec.push_back(pii(cnt / 2, t3));elsevec.push_back(pii(0, i));}}sort(vec.begin(), vec.end());int root = vec[vec.size() - 1].second;for(int i = vec.size() - 2; i >= 0; i--) {add_edge(root, vec[i].second);}Max_dis = -1;dfs(0, 0, 0);Max_dis = -1;dfs(key, key, 0);printf("%d\n", Max_dis);return 0;
}

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