环球旅行

原题地址：

https://ac.nowcoder.com/acm/problem/201723

基本题意：

切断树的一条边,将树分为两棵,要让两棵树的最长路径(即直径)中较大的那个最小。

基本思路：

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f
inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 1e6 + 10;
struct Edge{
    int to,val,next;
}edge[maxn << 1];
int n,cnt,head[maxn];
int l,r,p;
void add_edge(int u,int v,int w) {
  edge[++cnt].next = head[u];
  edge[cnt].to = v;
  edge[cnt].val = w;
  head[u] = cnt;
}
void dfs1(int u,int par,int sum,int flag) {//通过两次dfs1计算直径两端点;
  if(sum > p){
    p = sum;
    if(flag) r = u;
    else l = u;
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    if (to == par) continue;
    dfs1(to, u, sum + edge[i].val,flag);
  }
}
bool fd = false;
vector<pii> link;
void dfs3(int u,int par,int sum){//dfs3用来记录直径两个端点之间的路径;
  if(sum == p){
    fd = true;
    return;
  }
  for(int i = head[u] ; i != -1 ; i = edge[i].next){
    int to = edge[i].to;
    if(to == par) continue;
    link.emplace_back(mp(u,to));
    dfs3(to,u,sum + edge[i].val);
    if(fd) return;
    link.pop_back();
  }
}
void get_dia() {//直径端点预处理;
  p = 0;
  dfs1(1, -1, 0, 0);
  p = 0;
  dfs1(l, -1, 0, 1);
  dfs3(l, -1, 0);
}
int f[2][maxn],mx[2][maxn];//f记录子树的最长路径,mx记录子树的最长链;
void dfs2(int u,int par,int flag) {
  mx[flag][u] = f[flag][u] = 0;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    if (par == to) continue;
    dfs2(to, u, flag);
    //这里是在树形dp求直径里的经典技巧,具体可以百度搜索一下;
    f[flag][u] = max(f[flag][u],mx[flag][u] + mx[flag][to] + edge[i].val);
    mx[flag][u] = max(mx[flag][u], mx[flag][to] + edge[i].val);//以u为根的最长链;
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    f[flag][u] = max(f[flag][u], f[flag][to]);//要和子树中的最长路径比较更新;
  }
}
signed main() {
  cnt = 0;
  mset(head, -1);
  n = read();
  rep(i, 1, n - 1) {
    int u, v, w;
    u = read(), v = read(), w = read();
    add_edge(u, v, w);
    add_edge(v, u, w);
  }
  get_dia();
  dfs2(l, -1, 0);//处理出以直径左端点为根的每个子树的最长路径;
  dfs2(r, -1, 1);//同理处理出以直径右端点为根的每个子树的最长路径;
  int ans = INF;
  for (auto it : link) {
    int temp = max(f[1][it.first], f[0][it.second]); // 取左右两棵子树的最大值;
    ans = min(temp, ans);//要这个最大值最小;
  }
  print(ans);
  return 0;
}