P4180 严格次小生成树

https://www.luogu.org/problemnew/show/P4180

先吐槽一波垃圾蓝书，n^2logn有个矩阵用

先求出最小生成树，显然的，次小生成树应该是向MST中删一条树边加一条非树边

设加入的边连接u,v，为保证次小，应该删去MST中u->v路径上的边权最大值

考虑一个细节：如果加入的边和u->v上的边权最大值相同，此时应删去次大值

所以我们要完成这些任务：求树上两点之间边权的最大值、次大值

LCA即可

哦对inf要开大，因为tot最大可达10 ^ 14，开小了会WA掉50%...

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
using namespace std;
#define LM qwq
typedef int mainint;
#define int long long
inline int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (ans *= 10) += ch - '0';
        ch = getchar();
    }
    return ans * op;
}
#define inf 99999999999999
const int maxn = 1e5 + 5;
int n,m;
struct edge
{
    int to,next,cost;
}e[maxn * 6];
struct ed
{
    int u,v,w;
    bool operator < (const ed& b) const
    {
        return w < b.w;
    }
}a[maxn * 6];
bool vis[maxn * 6];
int fir[maxn],alloc;
void adde(int u,int v,int w)
{
    e[++alloc].next = fir[u];
    fir[u] = alloc;
    e[alloc].to = v;
    e[alloc].cost = w;
    swap(u,v);
    e[++alloc].next = fir[u];
    fir[u] = alloc;
    e[alloc].to = v;
    e[alloc].cost = w;    
}

int par[maxn];
int find(int x) { return x == par[x] ? x : par[x] = find(par[x]); }
void merge(int x,int y) { x = find(x),y = find(y); par[x] = y; }
bool same(int x,int y) { return find(x) == find(y); }
int tot;
void kruskal()
{
    for(int i = 1;i <= n;i++) par[i] = i;
    sort(a + 1,a + 1 + m);
    for(int i = 1;i <= m;i++)
    {
        int u = a[i].u,v = a[i].v;
        if(same(u,v)) continue;
        merge(u,v);
        adde(u,v,a[i].w);
        vis[i] = 1;
        tot += a[i].w;
    }
}

int dep[maxn],f[maxn][21],mx[maxn][21],mi[maxn][21];
void dfs(int u,int fa)
{
    dep[u] = dep[fa] + 1;
    for(int i = 1;i <= 20;i++)
    {
        f[u][i] = f[f[u][i - 1]][i - 1];
        mx[u][i] = max(mx[u][i - 1],mx[f[u][i - 1]][i - 1]);
        mi[u][i] = max(mi[u][i - 1],mi[f[u][i - 1]][i - 1]);
        if(mx[u][i - 1] > mx[f[u][i - 1]][i - 1]) mi[u][i] = max(mi[u][i],mx[f[u][i - 1]][i - 1]);
        else if(mx[u][i - 1] < mx[f[u][i - 1]][i - 1]) mi[u][i] = max(mi[u][i],mx[u][i - 1]);
    }
    for(int i = fir[u];i;i = e[i].next)
    {
        int v = e[i].to,w = e[i].cost;
        if(v == fa) continue;
        mx[v][0] = w;
        mi[v][0] = -inf;
        f[v][0] = u;
        dfs(v,u);
    }
}
int lca(int x,int y)
{
    if(dep[x] < dep[y]) swap(x,y);
    for(int i = 20;i >= 0;i--)
    {
        if(dep[f[x][i]] >= dep[y]) x = f[x][i];
        if(x == y) return x;
    }
    for(int i = 20;i >= 0;i--)
        if(f[x][i] != f[y][i]) x = f[x][i],y = f[y][i];
    return f[x][0];
}

int get_max(int u,int v,int val)
{
    int ans = -inf;
    for(int i = 20;i >= 0;i--)
    {
        if(dep[f[u][i]] >= dep[v])
        {
            if(mx[u][i] != val) ans = max(ans,mx[u][i]);
            else ans = max(ans,mi[u][i]);
            u = f[u][i];
        }
    }
    return ans;
}
mainint main()
{
    n = read(),m = read();
    for(int i = 1;i <= m;i++)
        a[i].u = read(),a[i].v = read(),a[i].w = read();
    kruskal();
    mi[1][0] = -inf;
    dfs(1,0);
    int ans = inf;
    for(int i = 1;i <= m;i++)
    {
        if(vis[i]) continue;
        int u = a[i].u,v = a[i].v,w = a[i].w;
        int fa = lca(u,v);
        //printf("%d %d %d\n",u,v,fa); 
        int maxm = max(get_max(u,fa,w),get_max(v,fa,w));
        //cout << maxm << endl;
        ans = min(ans,tot - maxm + w);
    }
    printf("%lld",ans);
}