最近公共祖先--------次小生成树

给定一张 N 个点 M 条边的无向图，求无向图的严格次小生成树。
设最小生成树的边权之和为sum，严格次小生成树就是指边权之和大于sum的生成树中最小的一个。
输入格式
第一行包含两个整数N和M。
接下来M行，每行包含三个整数x，y，z，表示点x和点y之前存在一条边，边的权值为z。
输出格式
包含一行，仅一个数，表示严格次小生成树的边权和。(数据保证必定存在严格次小生成树)
数据范围
N≤105,M≤3∗105N≤105,M≤3∗105
输入样例：
5 6
1 2 1
1 3 2
2 4 3
3 5 4
3 4 3
4 5 6

输出样例：
11

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010, M = 300010, INF = 0x3f3f3f3f;
int n, m;
typedef long long LL;
struct Edge {
 int a, b, w;
 bool used;
 bool operator < (const Edge& t) const{
  return w < t.w;
 }
}edge[M];
int p[N];
int h[N], e[M], w[M], ne[M], idx;
int depth[N], fa[N][17], d1[N][17], d2[N][17];
int q[N];
int find(int x) {
 if (p[x] != x)   p[x] = find(p[x]);
 return p[x];
}
void add(int a, int b, int c) {
 e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
void build() {
 memset(h, -1, sizeof h);
 for (int i = 0; i < m; i ++)
  if (edge[i].used) {
   int a = edge[i].a, b = edge[i].b, w = edge[i].w;
   add(a, b, w), add(b, a, w);
  }
}
LL kruskal() {
 for (int i = 1; i <= n; i++)    p[i] = i;
 sort(edge, edge + m);
 LL res = 0;
for (int i = 0; i < m; i++) {
  int a = find(edge[i].a), b = find(edge[i].b), w = edge[i].w;
  if (a != b) {
   p[a] = b;
   res += w;
   edge[i].used = true;
  }
 }
 return res;
}
void bfs() {
 memset(depth, 0x3f, sizeof depth);
 depth[0] = 0, depth[1] = 1;
 q[0] = 1;
int hh = 0, tt = 0;
 while (hh <= tt) {
  int t = q[hh++];
  for (int i = h[t]; ~i; i = ne[i]) {
   int j = e[i];
   if (depth[j] > depth[t] + 1) {
    depth[j] = depth[t] + 1;
    q[++tt] = j;
    fa[j][0] = t;
    d1[j][0] = w[i], d2[j][0] = -INF;
    for (int k = 1; k <= 16; k++) {
     int anc = fa[j][k - 1];
     fa[j][k] = fa[anc][k - 1];
     int distance[4] = { d1[j][k - 1], d2[j][k - 1], d1[anc][k - 1], d2[anc][k - 1] };
     d1[j][k] = d2[j][k] = -INF;
     for (int u = 0; u < 4; u++) {
      int d = distance[u];
      if (d > d1[j][k])    d2[j][k] = d1[j][k], d1[j][k] = d;
      else                 if (d != d1[j][k] && d > d2[j][k])     d2[j][k] = d;
     }
    }
   }
  }
 }
}
int lca(int a, int b, int w) {
 static int distance[N * 2];
 int cnt = 0;
 if (depth[a] < depth[b])   swap(a, b);
 for (int k = 16; k >= 0 ; k --)
  if (depth[fa[a][k]] >= depth[b]) {
   distance[cnt ++] = d1[a][k];
   distance[cnt++] = d2[a][k];
   a = fa[a][k];
  }
 if (a != b) {
  for (int k = 16; k >= 0; k--)
   if (fa[a][k] != fa[b][k]) {
    distance[cnt++] = d1[a][k];
    distance[cnt++] = d2[a][k];
    distance[cnt++] = d1[b][k];
    distance[cnt++] = d2[b][k];
    a = fa[a][k], b = fa[b][k];
   }
  distance[cnt++] = d1[a][0];
  distance[cnt++] = d1[b][0];
 }
 int dist1 = -INF, dist2 = -INF;
 for (int i = 0; i < cnt; i++) {
  int d = distance[i];
  if (d > dist1)   dist2 = dist1, dist1 = d;
  else    if (d != dist1 && d > dist2)   dist2 = d;
 }
 if (w > dist1)    return w - dist1;
 if (w > dist2)    return w - dist2;
 return INF;
 }
int main() {
 cin >> n >> m;
 for (int i = 0; i < m; i++) {
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  edge[i] = { a, b, c };
 }
 LL sum = kruskal();
 build();
 bfs();
 LL res = 1e18;
 for (int i = 0; i < m; i++) {
  if (!edge[i].used) {
   int a = edge[i].a, b = edge[i].b, w = edge[i].w;
   res = min(res, sum + lca(a, b, w));
  }
 }
 printf("%lld\n", res);
 return 0;
}