zz:https://www.cnblogs.com/ytytzzz/p/9674661.html
题意：
给一棵树，每次询问删掉两条边，问剩下的三棵树的最大直径
点10W，询问10W，询问相互独立
Solution：
考虑线段树/倍增维护树的直径
考虑一个点集的区间 [l, r]
而我们知道了有 l <= k < r，
且知道 [l, k] 和 [k + 1, r] 两个区间的最长链的端点及长度
假设两个区间的直径端点分别为 (l1, r1) 和 (l2, r2)
那么 [l, r] 这个区间的直径长度为
dis(l1, r1) dis(l1, l1) dis(l1, r2)
dis(r1, l2) dis(r1, r2) dis(l2, r2)
六个值中的最大值
本题因为操作子树，所以我们维护dfs序的区间最长链即可
#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = 2e5 + 5;
int T, n, m;
int len, head[N], ST[20][N];
struct edge{int u, v, w;}ee[N];
int cnt, fa[N], log_2[N], st[N], en[N], dfn[N], dis[N], dep[N], pos[N];
struct edges{int to, next, cost;}e[N];
inline void add(int u, int v, int w) {
e[++ len] = (edges){v, head[u], w}, head[u] = len;
e[++ len] = (edges){u, head[v], w}, head[v] = len;
}
inline void dfs1(int u) {
st[u] = ++ cnt, dfn[cnt] = u;
for (int v, i = head[u]; i; i = e[i].next) {
v = e[i].to;
if (v == fa[u]) continue;
fa[v] = u, dep[v] = dep[u] + 1;
dis[v] = dis[u] + e[i].cost, dfs1(v);
}
en[u] = cnt;
}
inline void dfs2(int u) {
dfn[++ cnt] = u, pos[u] = cnt;
for (int v, i = head[u]; i; i = e[i].next) {
v = e[i].to;
if (v == fa[u]) continue;
dfs2(v), dfn[++ cnt] = u;
}
}
int mmin(int x, int y) {
if (dep[x] < dep[y]) return x;
return y;
}
inline int lca(int u, int v) {
static int w;
if (pos[u] > pos[v]) swap(u, v);
w = log_2[pos[v] - pos[u] + 1];
return mmin(ST[w][pos[u]], ST[w][pos[v] - (1 << w) + 1]);
}
inline int dist(int u, int v) {
int Lca = lca(u, v);
return dis[u] + dis[v] - dis[Lca] * 2;
}
inline void build() {
for (int i = 1; i <= cnt; i ++)
ST[0][i] = dfn[i];
for (int i = 1; i < 20; i ++)
for (int j = 1; j <= cnt; j ++)
if (j + (1 << (i - 1)) > cnt) ST[i][j] = ST[i - 1][j];
else ST[i][j] = mmin(ST[i - 1][j], ST[i - 1][j + (1 << (i - 1))]);
}
int M;
struct node {
int l, r, dis;
}tr[N << 1];
inline void update(int o, int o1, int o2) {
static int d;
static node tmp;
if (tr[o1].dis == -1) {tr[o] = tr[o2]; return;}
if (tr[o2].dis == -1) {tr[o] = tr[o1]; return;}
if (tr[o1].dis > tr[o2].dis) tmp = tr[o1];
else tmp = tr[o2];
d = dist(tr[o1].l, tr[o2].l);
if (d > tmp.dis) tmp.l = tr[o1].l, tmp.r = tr[o2].l, tmp.dis = d;
d = dist(tr[o1].l, tr[o2].r);
if (d > tmp.dis) tmp.l = tr[o1].l, tmp.r = tr[o2].r, tmp.dis = d;
d = dist(tr[o1].r, tr[o2].l);
if (d > tmp.dis) tmp.l = tr[o1].r, tmp.r = tr[o2].l, tmp.dis = d;
d = dist(tr[o1].r, tr[o2].r);
if (d > tmp.dis) tmp.l = tr[o1].r, tmp.r = tr[o2].r, tmp.dis = d;
tr[o] = tmp;
}
inline void ask(int s, int t) {
if (s > t) return;
for (s += M - 1, t += M + 1; s ^ t ^ 1; s >>= 1, t >>= 1) {
if (~s&1) update(0, 0, s ^ 1);
if ( t&1) update(0, 0, t ^ 1);
}
}
inline int get_char() {
static const int SIZE = 1 << 23;
static char *T, *S = T, buf[SIZE];
if (S == T) {
T = fread(buf, 1, SIZE, stdin) + (S = buf);
if (S == T) return -1;
}
return *S ++;
}
inline void in(int &x) {
static int ch;
while (ch = get_char(), ch > 57 || ch < 48);x = ch - 48;
while (ch = get_char(), ch > 47 && ch < 58) x = x * 10 + ch - 48;
}
int main() {
int u, v, w, ans;
log_2[1] = 0;
for (int i = 2; i <= 200000; i ++)
if (i == 1 << (log_2[i - 1] + 1))
log_2[i] = log_2[i - 1] + 1;
else log_2[i] = log_2[i - 1];
for (in(T); T --; ) {
in(n), in(m), cnt = len = 0;
for (int i = 1; i <= n; i ++)
head[i] = 0;
for (int i = 1; i < n; i ++) {
in(ee[i].u), in(ee[i].v), in(ee[i].w);
add(ee[i].u, ee[i].v, ee[i].w);
}
dfs1(1);
for (M = 1; M < n + 2; M <<= 1);
for (int i = 1; i <= n; i ++)
tr[i + M].l = tr[i + M].r = dfn[i], tr[i + M].dis = 0;
for (int i = n + M + 1; i <= (M << 1) + 1; i ++)
tr[i].dis = -1;
cnt = 0, dfs2(1), build();
for (int i = M; i; i --)
update(i, i << 1, i << 1 | 1);
for (int i = 1; i < n; i ++)
if (dep[ee[i].u] > dep[ee[i].v])
swap(ee[i].u, ee[i].v);
for (int u, v, i = 1; i <= m; i ++) {
in(u), in(v), ans = 0;
u = ee[u].v, v = ee[v].v, w = lca(u, v);
if (w == u || w == v) {
if (w != u) swap(u, v);
tr[0].dis = -1, ask(1, st[u] - 1), ask(en[u] + 1, n), ans = max(ans, tr[0].dis);
tr[0].dis = -1, ask(st[u], st[v] - 1), ask(en[v] + 1, en[u]), ans = max(ans, tr[0].dis);
tr[0].dis = -1, ask(st[v], en[v]), ans = max(ans, tr[0].dis);
}
else {
if (st[u] > st[v]) swap(u, v);
tr[0].dis = -1, ask(1, st[u] - 1), ask(en[u] + 1, st[v] - 1), ask(en[v] + 1, n), ans = max(ans, tr[0].dis);
tr[0].dis = -1, ask(st[u], en[u]), ans = max(ans, tr[0].dis);
tr[0].dis = -1, ask(st[v], en[v]), ans = max(ans, tr[0].dis);
}
printf("%d\n", ans);
}
}
return 0;
}