Brief Description

您有一个无向带权图，您需要支持两种操作。

1. 询问两个点之间的最大权最小路径。
2. 删除一条边。

Algorithm Design

我们首先提出一个猜想：最优路径一定在原图的一个最小生成森林上，证明如下：
假设最优路径有\(\phi\)条边不再最小生成森林上，我们考察其中的一条边，根据定义，生成森林中一定有一条路径链接这条边的顶点且权值和小于这条边，那么我们如果使用这条路径代替这条边，\(\phi\)会减小而最大权不会增加，所以根据反证法我们就可以知道这个结论的正确性。
有了这个引理之后我们可以把原题转化为维护一个最小生成森林，这显然可以使用lct维护。
考虑细节。因为lct不是太支持删一条边，所以我们离线地倒过来做。
另外我们需要记录每条边是否被删。我开始使用了map，然而不知道为什么一直RE。后来改用二分查找就好了。所以——

抵制STL从我做起！（逃

Code

#include <algorithm>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
const int maxv = 1500005;
int read() {
int x = 0, f = 1;
char ch = getchar();
while (!isdigit(ch)) {
if (ch == '-')
f = -1;
ch = getchar();
}
while (isdigit(ch)) {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
struct data {
int from, to, dat, br, id;
data(int x = 0, int y = 0, int z = 0, int i = 0, int j = 0)
: from(x), to(y), dat(z), br(i), id(j) {}
bool operator<(const data b) const {
return this->from < b.from || ((this->from == b.from) && (this->to < b.to));
}
} a[1000005];
struct req {
int opt, x, y, id, ans;
req(int x = 0, int y = 0, int z = 0, int k = 0, int fuck = 0) {
this->opt = x;
this->x = y;
this->y = z;
this->id = k;
this->ans = fuck;
}
} q[100005];
bool cmp(data a, data b) { return a.dat < b.dat; }
bool cmp2(data a, data b) { return a.id < b.id; }
int n, m, qaq, f[maxv], max[maxv], val[maxv], tot, maxnum[maxv];
int fa[maxv], ch[maxv][2];
bool rev[maxv];
int bisearch(int u, int v) {
int l = 1, r = m;
while (l <= r) {
int mid = (l + r) >> 1;
if (a[mid].from < u || (a[mid].from == u && a[mid].to < v))
l = mid + 1;
else if (a[mid].from == u && a[mid].to == v)
return mid;
else
r = mid - 1;
}
return -1;
}
bool isroot(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
void pushdown(int k) {
if (rev[k]) {
rev[k] ^= 1;
rev[ch[k][0]] ^= 1;
rev[ch[k][1]] ^= 1;
std::swap(ch[k][0], ch[k][1]);
}
}
void update(int x) {
maxnum[x] = x;
int l = maxnum[ch[x][0]], r = maxnum[ch[x][1]];
if (val[l] > val[maxnum[x]])
maxnum[x] = l;
if (val[r] > val[maxnum[x]])
maxnum[x] = r;
max[x] = val[maxnum[x]];
}
void zig(int x) {
int y = fa[x], z = fa[y], l = (ch[y][1] == x), r = l ^ 1;
if (!isroot(y))
ch[z][ch[z][1] == y] = x;
fa[ch[y][l] = ch[x][r]] = y;
fa[ch[x][r] = y] = x;
fa[x] = z;
update(y);
update(x);
}
void splay(int x) {
int s[maxv], top = 0;
s[++top] = x;
for (int i = x; !isroot(i); i = fa[i])
s[++top] = fa[i];
while (top)
pushdown(s[top--]);
for (int y; !isroot(x); zig(x)) {
if (!isroot(y = fa[x])) {
zig((ch[fa[y]][0] == y) == (ch[y][0] == x) ? y : x);
}
}
update(x);
}
void access(int x) {
for (int t = 0; x; t = x, x = fa[x]) {
splay(x);
ch[x][1] = t;
update(x);
}
}
void makeroot(int x) {
access(x);
splay(x);
rev[x] ^= 1;
}
void split(int x, int y) {
makeroot(y);
access(x);
splay(x);
}
void link(int x, int y) {
makeroot(x);
fa[x] = y;
}
void cut(int x, int y) {
makeroot(x);
access(y);
splay(y);
ch[y][0] = fa[x] = 0;
}
void init() {
memset(val, 0, sizeof(val));
n = read();
m = read();
qaq = read();
for (int i = 1; i <= n; i++)
f[i] = i;
for (int i = 1; i <= m; i++) {
int x = read(), y = read(), z = read();
if (x > y)
std::swap(x, y);
a[i] = data(x, y, z);
}
std::sort(a + 1, a + 1 + m, cmp);
for (int i = 1; i <= m; i++) {
a[i].id = i;
val[n + i] = a[i].dat;
maxnum[n + i] = n + i;
}
std::sort(a + 1, a + 1 + m);
for (int i = 1; i <= qaq; i++) {
int x = read(), y = read(), z = read();
q[i] = req(x, y, z);
if (x == 2) {
if (q[i].x > q[i].y)
std::swap(q[i].x, q[i].y);
int t = bisearch(q[i].x, q[i].y);
a[t].br = 1;
q[i].id = a[t].id;
}
}
}
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
void kruskal() {
std::sort(a + 1, a + m + 1, cmp2);
tot = 0;
for (int i = 1; i <= m; i++) {
if (!a[i].br) {
int u = a[i].from, v = a[i].to, x = find(u), y = find(v);
if (x != y) {
f[x] = y;
link(u, i + n);
link(v, i + n);
tot++;
if (tot == n - 1)
break;
}
}
}
}
void solve() {
for (int i = qaq; i >= 1; i--) {
int op = q[i].opt, x = q[i].x, y = q[i].y;
if (op == 1) {
split(x, y);
q[i].ans = val[maxnum[x]];
}
if (op == 2) {
int k = q[i].id;
split(x, y);
int t = maxnum[x];
if (a[k].dat < val[t]) {
cut(a[t - n].from, t);
cut(a[t - n].to, t);
link(x, k + n);
link(y, k + n);
}
}
}
for (int i = 1; i <= qaq; i++) {
if (q[i].opt == 1)
printf("%d\n", q[i].ans);
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input", "r", stdin);
#endif
init();
kruskal();
solve();
return 0;
}