题意：有一棵树，n个点，n-1条边每条边有一个权值，有三种操作，query a b，询问a到b所有路径的最大权值，negate a b，把a到b所有路径权值设置为对应权值的相反数，change a b，把第a条边权值设为b。
题解：线段树维护最大值和最小值，negate a b的时候把对应区间最大最小值交换并取反，注意要用一个标记每个区间取反的次数，因为偶数次取反相当于不变。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 10005;
const int INF = 0x3f3f3f3f;
struct Edge {
int u, v, w, nxt;
Edge() {}
Edge(int a, int b, int c, int d): u(a), v(b), w(c), nxt(d) {}
}e[N << 1];
int size[N], son[N], fa[N], top[N], id[N], dep[N];
int n, cnt, tot, head[N], minn[N << 2], maxx[N << 2], flag[N << 2];
void AddEdge(int u, int v, int w) {
e[cnt] = Edge(u, v, w, head[u]);
head[u] = cnt++;
e[cnt] = Edge(v, u, w, head[v]);
head[v] = cnt++;
}
void dfs2(int u, int tp) {
top[u] = tp;
id[u] = ++tot;
if (son[u]) dfs2(son[u], tp);
for (int i = head[u]; i + 1; i = e[i].nxt) {
int v = e[i].v;
if (v == fa[u] || v == son[u]) continue;
dfs2(v, v);
}
}
void dfs1(int u, int f, int depth) {
size[u] = 1, son[u] = 0, dep[u] = depth, fa[u] = f;
for (int i = head[u]; i + 1; i = e[i].nxt) {
int v = e[i].v;
if (v == f) continue;
dfs1(v, u, depth + 1);
size[u] += size[v];
if (size[v] > size[son[u]]) son[u] = v;
}
}
void pushup(int k) {
minn[k] = min(minn[k * 2], minn[k * 2 + 1]);
maxx[k] = max(maxx[k * 2], maxx[k * 2 + 1]);
}
void pushdown(int k) {
if (flag[k]) {
flag[k * 2] ^= 1;
flag[k * 2 + 1] ^= 1;
swap(minn[k * 2], maxx[k * 2]);
minn[k * 2] = -minn[k * 2];
maxx[k * 2] = -maxx[k * 2];
swap(minn[k * 2 + 1], maxx[k * 2 + 1]);
minn[k * 2 + 1] = -minn[k * 2 + 1];
maxx[k * 2 + 1] = -maxx[k * 2 + 1];
flag[k] = 0;
}
}
void modify(int k, int left, int right, int pos, int v) {
if (left == right) {
minn[k] = maxx[k] = v;
return;
}
pushdown(k);
int mid = (left + right) / 2;
if (pos <= mid)
modify(k * 2, left, mid, pos, v);
else
modify(k * 2 + 1, mid + 1, right, pos, v);
pushup(k);
}
void negate(int k, int left, int right, int l, int r) {
if (l <= left && right <= r) {
swap(minn[k], maxx[k]);
minn[k] = -minn[k];
maxx[k] = -maxx[k];
flag[k] ^= 1;
return;
}
pushdown(k);
int mid = (left + right) / 2;
if (l <= mid)
negate(k * 2, left, mid, l, r);
if (r > mid)
negate(k * 2 + 1, mid + 1, right, l, r);
pushup(k);
}
int query(int k, int left, int right, int l, int r) {
if (l <= left && right <= r)
return maxx[k];
pushdown(k);
int mid = (left + right) / 2, res = -INF;
if (l <= mid)
res = max(res, query(k * 2, left, mid, l, r));
if (r > mid)
res = max(res, query(k * 2 + 1, mid + 1, right, l, r));
return res;
}
void solve(int u, int v, int flag) {
int tp1 = top[u], tp2 = top[v], res = -INF;
while (tp1 != tp2) {
if (dep[tp1] < dep[tp2]) {
swap(tp1, tp2);
swap(u, v);
}
if (!flag) res = max(res, query(1, 1, tot, id[tp1], id[u]));
else negate(1, 1, tot, id[tp1], id[u]);
u = fa[tp1];
tp1 = top[u];
}
if (dep[u] > dep[v]) swap(u, v);
if (!flag) res = max(res, query(1, 1, tot, id[son[u]], id[v]));
else negate(1, 1, tot, id[son[u]], id[v]);
if (!flag) printf("%d\n", res);
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
int u, v, w;
memset(head, -1, sizeof(head));
memset(minn, INF, sizeof(minn));
memset(maxx, -INF, sizeof(maxx));
memset(flag, 0, sizeof(flag));
cnt = tot = 0;
scanf("%d", &n);
for (int i = 0; i < n - 1; i++) {
scanf("%d%d%d", &u, &v, &w);
AddEdge(u, v, w);
}
dfs1(1, 0, 1);
dfs2(1, 1);
for (int i = 0; i < cnt; i += 2) {
u = e[i].u, v = e[i].v;
if (dep[u] < dep[v]) swap(u, v);
modify(1, 1, tot, id[u], e[i].w);
}
char op[10];
int a, b;
while (scanf("%s", op) && op[0] != 'D') {
scanf("%d%d", &a, &b);
if (op[0] == 'Q')
solve(a, b, 0);
else if (op[0] == 'N')
solve(a, b, 1);
else {
u = e[a - 1 << 1].u;
v = e[a - 1 << 1].v;
if (dep[u] < dep[v]) swap(u, v);
modify(1, 1, tot, id[u], b);
}
}
}
return 0;
}
