【题解】[WC2006]水管局长

感觉这题好强啊……本来以为能过，结果毫无疑问的被ge了一顿……在这里记录一下做的过程，也免得以后又忘记啦。

首先，我们应看出在这张图上，要让经过的水管最长的最短，就是要维护一棵动态的最小生成树。只是删除边不是很好操作，所以我们将删边改为加边，反向处理。如果发现新加的边可以更新最小生成树，我们就应该更新图，删去原图中最大的一条边。所以在LCT上我们要多维护一个信息，即当前最大的那条边是哪一条边。至此，大部分的难点都已解决，只是LCT维护的是点上的信息，而这道题的权值都在边上。我们将边转化成点，要将要链接的两个点分别与另一点相连，将信息存在另一个点上即可。

The road is tough and there's still a long way to go. Try your best and don't regret.

#include <bits/stdc++.h>
using namespace std;
#define maxn 150005
#define maxm 1500000
int n, m, opt[maxm], cnp, x[maxm], y[maxm];
int q[maxm], id, T[maxm];
int tot, ans[maxn], mark[maxn];
map <int, int> Map[maxn];
struct node
{
    int fa, sum, rev;
    int id, v, ch[2];
}P[maxm];

struct edge
{
    int u, v, t;
    friend bool operator <(edge a, edge b)
    {
        return a.t < b.t;
    }
}E[maxn];

int read()
{
    int x = 0, k = 1;
    char c;
    c = getchar();
    while(c < '0' || c > '9') { if(c == '-') k = -1; c = getchar(); }
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * k;
}

void push_down(int x) 
{
    if(!P[x].rev) return;
    swap(P[x].ch[0], P[x].ch[1]);
    P[P[x].ch[0]].rev ^= 1, P[P[x].ch[1]].rev ^= 1;
    P[x].rev = 0;
}

void update(int x)
{
    if(!x) return;
    P[x].id = x, P[x].sum = P[x].v;
    if(P[P[x].ch[0]].sum > P[x].sum) P[x].sum = P[P[x].ch[0]].sum, P[x].id = P[P[x].ch[0]].id;
    if(P[P[x].ch[1]].sum > P[x].sum) P[x].sum = P[P[x].ch[1]].sum, P[x].id = P[P[x].ch[1]].id;
}

bool is_root(int x) { return (P[P[x].fa].ch[0] != x) && (P[P[x].fa].ch[1] != x); }

void rotate(int x)
{
    int f = P[x].fa, gf = P[f].fa, k = P[f].ch[1] == x;
    if(!is_root(f)) P[gf].ch[P[gf].ch[1] == f] = x;
    P[x].fa = gf, P[f].fa = x;
    P[f].ch[k] = P[x].ch[k ^ 1], P[P[x].ch[k ^ 1]].fa = f;
    P[x].ch[k ^ 1] = f;
    update(f), update(x);
}

void Splay(int x)
{
    int top = 1; q[top] = x;
    for(int i = x; !is_root(i); i = P[i].fa) q[++ top] = P[i].fa;
    for(int i = top; i; i --) push_down(q[i]);
    while(!is_root(x))
    {
        int f = P[x].fa, gf = P[f].fa;
        if(!is_root(f)) (P[f].ch[1] == x) ^ (P[gf].ch[1] == f) ? rotate(x) : rotate(f);
        rotate(x);
    }
    update(x);
}

void Access(int x)
{
    for(int ff = 0; x; ff = x, x = P[x].fa)
    {
        Splay(x);
        P[x].ch[1] = ff; 
        update(x);
    }
}

void make_root(int x) {  Access(x); Splay(x); P[x].rev ^= 1; }
void Split(int u, int v) { make_root(u); Access(v); Splay(v); }
void Link(int u, int v) { make_root(u); P[u].fa = v; }
void Cut(int u, int v) { Split(u, v); P[v].ch[0] = P[u].fa = 0; }

int Get_fa(int x)
{
    Access(x); Splay(x);
    while(P[x].ch[0]) x = P[x].ch[0];
    return x;
}

int main()
{
    n = read(), m = read();
    int q = read();
    for(int i = 1; i <= m; i ++) 
    {
        int u = read(), v = read(), w = read(); T[i] = w;
        E[++ cnp].u = u; E[cnp].v = v, E[cnp].t = w;
    }
    sort(E + 1, E + 1 + m);
    for(int i = 1; i <= m; i ++) Map[E[i].u][E[i].v] = Map[E[i].v][E[i].u] = i;
    for(int i = 1; i <= q; i ++) 
    {
        opt[i] = read(); x[i] = read(); y[i] = read();
        if(opt[i] == 2) mark[Map[x[i]][y[i]]] = 1;
    }
    for(int i = 1; i <= m; i ++)
    {
        int u = E[i].u, v = E[i].v, t = E[i].t;
        if(!mark[Map[u][v]] && Get_fa(u) != Get_fa(v)) 
        {
            id = i + n, P[id].sum = P[id].v = t;
            Link(u, id), Link(id, v);
        }
     }
     for(int i = q; i; i --)
    {
        Split(x[i], y[i]);
        if(opt[i] == 1) // find a route
            ans[++ tot] = P[y[i]].sum;
        else // add edge
        {
            int now = n + Map[x[i]][y[i]];
            int tem = E[now - n].t;
            if(tem < P[y[i]].sum)
            {
                P[now].v = tem; tem = P[y[i]].id;
                Cut(tem, E[tem - n].u); Cut(tem, E[tem - n].v);
                Link(now, x[i]), Link(now, y[i]);
            }
        }
    }
    for(int i = tot; i; i --)
        printf("%d
", ans[i]);
    return 0;
}