Description：

给一个图，求他的严格次小生成树。

$1\leqslant n\leqslant 10^5$

Solution：

先用 $Kruskal$ 求出来一个生成树，然后构建LCT，维护这条链的最大值和次大值，那么在加入一条非树边时会产生一个环，并且这条边一定大于环上所有边，于是如果边长不等于最大值，就用它和最大值的差更新答案，否则用和次大值的，也就是说最大值和次大值必不相同。

Code：

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cmath>

#include<stack>

#include<cctype>

#include<cstring>

using namespace std;

inline int rd()

{

	register int res = 0,f = 1;

	register char c = getchar();

	while(!isdigit(c))

	{

		if(c == '-')f = -1;

		c = getchar();

	}

	while(isdigit(c))

	{

		res = (res << 1) + (res << 3) + c - '0';

		c = getchar();

	}

	return res * f;

}

int n,m;

#define MAXN 100010

#define MAXM 300010

struct edges

{

	int u,v,w;

	bool tag;

}es[MAXM];

bool cmp_w(edges a,edges b){return a.w < b.w;}

struct node

{

	int c[2],fa,val,mv[2];

	bool rev;

}t[MAXN << 1];

inline bool isroot(int k){return (t[t[k].fa].c[0] != k && t[t[k].fa].c[1] != k);}

inline void connect(int k,int f,int p){t[k].fa = f;t[f].c[p] = k;return;}

inline int id(int k){return (t[t[k].fa].c[0] == k ? 0 : 1);}

inline void checkmax(int &a,int &b,int c,int d)

{

	if(c > a)

	{

		if(d > a)a = c,b = d;

		else b = a,a = c;

		return;

	}

	if(c == a){if(d > b)b = d;return;}

	if(c < a){if(c > b)b = c;return;}

}

inline void maintain(int k)

{

	t[k].mv[0] = t[k].val;t[k].mv[1] = 0xc0c0c0c0;

	if(t[k].c[0] != 0)checkmax(t[k].mv[0],t[k].mv[1],t[t[k].c[0]].mv[0],t[t[k].c[0]].mv[1]);

	if(t[k].c[1] != 0)checkmax(t[k].mv[0],t[k].mv[1],t[t[k].c[1]].mv[0],t[t[k].c[1]].mv[1]);

	return;

}

inline void pushdown(int k)

{

	if(t[k].rev)

	{

		t[t[k].c[0]].rev ^= 1;t[t[k].c[1]].rev ^= 1;

		swap(t[k].c[0],t[k].c[1]);t[k].rev = false;

	}

	return;

}

inline void rotate(int x)

{

	register int y = t[x].fa,z = t[y].fa,fx = id(x),fy = id(y);

	if(!isroot(y))t[z].c[fy] = x;

	t[x].fa = z;

	connect(t[x].c[fx ^ 1],y,fx);

	connect(y,x,fx ^ 1);

	maintain(y);maintain(x);

	return;

}

stack<int> s;

inline void splay(int x)

{

	s.push(x);

	for(register int i = x;!isroot(i);i = t[i].fa)s.push(t[i].fa);

	while(!s.empty()){pushdown(s.top());s.pop();}

	while(!isroot(x))

	{

		register int y = t[x].fa;

		if(isroot(y)){rotate(x);break;}

		if(id(x) == id(y)){rotate(y);rotate(x);}

		else{rotate(x);rotate(x);}

	}

	return;

}

inline void access(int k){for(register int i = 0;k;i = k,k = t[k].fa){splay(k);t[k].c[1] = i;maintain(k);}return;}

inline void makeroot(int k){access(k);splay(k);t[k].rev ^= 1;return;}

int f[MAXN];

int find(int x){return (x == f[x] ? x : f[x] = find(f[x]));}

inline void link(int x,int y){makeroot(x);t[x].fa = y;return;}

inline void split(int x,int y){makeroot(x);access(y);splay(y);return;}

int main()

{

	scanf("%d%d",&n,&m);

	for(register int i = 1;i <= m;++i)

	{

		es[i].u = rd();es[i].v = rd();

		es[i].w = rd();

	}

	for(register int i = 1;i <= n;++i)f[i] = i;

	sort(es + 1,es + 1 + m,cmp_w);

	register long long ans = 0x3f3f3f3f3f3f3f3f,sum = 0;

	register int tot = n;

	for(register int i = 1;i <= m;++i)

	{

		if(find(es[i].u) != find(es[i].v))

		{

			f[find(es[i].u)] = find(es[i].v);

			++tot;

			t[tot].val = es[i].w;

			link(es[i].u,tot);link(es[i].v,tot);

			maintain(tot);

			es[i].tag = true;

			sum += es[i].w;

		}

	}

	for(register int i = 1;i <= m;++i)

	{

		if(!es[i].tag)

		{

			split(es[i].u,es[i].v);

			if(es[i].w != t[es[i].v].mv[0])

			{

				ans = min(ans,sum + es[i].w - t[es[i].v].mv[0]);

			}

			else

			{

				ans = min(ans,sum + es[i].w - t[es[i].v].mv[1]);

			}

		}

	}

	cout << ans << endl;

	return 0;

}