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题意:3种操作,1单点更新,2路径正负反转,3路径查询最大值

思路:线段树维护最大值和最小值和一个懒惰标记,然后在线段树的基础上用树链剖分维护

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

const int MX = 1e4 + 5;
const int INF = 0x3f3f3f3f;

struct Edge {
int u, v, nxt, cost;
} E[MX << 2];
int Head[MX], h_r;
void edge_init() {
h_r = 0;
memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v, int cost) {
E[h_r].v = v; E[h_r].u = u; E[h_r].cost = cost;
E[h_r].nxt = Head[u];
Head[u] = h_r++;
}

int MAX[MX << 2], MIN[MX << 2], col[MX << 2], A[MX];
void push_up(int rt) {
MAX[rt] = max(MAX[rt << 1], MAX[rt << 1 | 1]);
MIN[rt] = min(MIN[rt << 1], MIN[rt << 1 | 1]);
}
void change(int rt) {
swap(MAX[rt], MIN[rt]);
MAX[rt] = -MAX[rt];
MIN[rt] = -MIN[rt];
}
void push_down(int rt) {
if(col[rt]) {
col[rt << 1] ^= 1;
col[rt << 1 | 1] ^= 1;
change(rt << 1); change(rt << 1 | 1);
col[rt] = 0;
}
}
void build(int l, int r, int rt) {
col[rt] = 0;
if(l == r) {
MAX[rt] = MIN[rt] = A[l];
return;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
push_up(rt);
}
void update_seg(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
change(rt);
col[rt] ^= 1;
return;
}
int m = (l + r) >> 1;
push_down(rt);
if(L <= m) update_seg(L, R, lson);
if(R > m) update_seg(L, R, rson);
push_up(rt);
}
void update(int x, int d, int l, int r, int rt) {
if(l == r) {
MAX[rt] = MIN[rt] = d;
return;
}
int m = (l + r) >> 1;
push_down(rt);
if(x <= m) update(x, d, lson);
else update(x, d, rson);
push_up(rt);
}
int query(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) {
return MAX[rt];
}
int m = (l + r) >> 1, ret = -INF;
push_down(rt);
if(L <= m) ret = max(ret, query(L, R, lson));
if(R > m) ret = max(ret, query(L, R, rson));
return ret;
}

int fa[MX], top[MX], siz[MX], son[MX], dep[MX], id[MX], rear;
void DFS1(int u, int f, int d) {
fa[u] = f; dep[u] = d;
son[u] = 0; siz[u] = 1;
for(int i = Head[u]; ~i; i = E[i].nxt) {
int v = E[i].v;
if(v == f) continue;
DFS1(v, u, d + 1);
siz[u] += siz[v];
if(siz[son[u]] < siz[v]) {
son[u] = v;
}
}
}
void DFS2(int u, int tp) {
top[u] = tp;
id[u] = ++rear;
if(son[u]) DFS2(son[u], tp);
for(int i = Head[u]; ~i; i = E[i].nxt) {
int v = E[i].v;
if(v == fa[u] || v == son[u]) continue;
DFS2(v, v);
}
}
/*节点1不使用,建树要小心
一般边使用更深的那个点的id编号来表示
*/
void HLD_presolve() {
rear = 0;
DFS1(1, 0, 1);
DFS2(1, 1);
for(int i = 0; i < 2 * (rear - 1); i += 2) {
int u = E[i].u, v = E[i].v;
if(dep[u] < dep[v]) swap(u, v);
A[id[u]] = E[i].cost;
}
A[1] = -INF;
build(1, rear, 1);
}
/*找到对应边的更深的点的id编号*/
void HLD_update(int x, int d) {
x = (x - 1) * 2;
int u = E[x].u, v = E[x].v;
if(dep[u] < dep[v]) swap(u, v);
update(id[u], d, 1, rear, 1);
}
/*注意最后一个查询与单点更新的区别以及u==v就需要返回x*/
void HLD_negate(int u, int v) {
int tp1 = top[u], tp2 = top[v];
while(tp1 != tp2) {
if(dep[tp1] < dep[tp2]) {
swap(u, v);
swap(tp1, tp2);
}
update_seg(id[tp1], id[u], 1, rear, 1);
u = fa[tp1]; tp1 = top[u];
}
if(u == v) return;
if(dep[u] > dep[v]) swap(u, v);
update_seg(id[son[u]], id[v], 1, rear, 1);
}
int HLD_query(int u, int v) {
int tp1 = top[u], tp2 = top[v], ans = -INF;
while(tp1 != tp2) {
if(dep[tp1] < dep[tp2]) {
swap(u, v);
swap(tp1, tp2);
}
ans = max(ans, query(id[tp1], id[u], 1, rear, 1));
u = fa[tp1]; tp1 = top[u];
}
if(u == v) return ans;
if(dep[u] > dep[v]) swap(u, v);
ans = max(ans, query(id[son[u]], id[v], 1, rear, 1));
return ans;
}

int main() {
int T, n; //FIN;
scanf("%d", &T);
while(T--) {
edge_init();
scanf("%d", &n);
for(int i = 1; i <= n - 1; i++) {
int u, v, cost;
scanf("%d%d%d", &u, &v, &cost);
edge_add(u, v, cost);
edge_add(v, u, cost);
}
HLD_presolve();

char op[10]; int a, b;
while(scanf("%s", op), op[0] != 'D') {
scanf("%d%d", &a, &b);
if(op[0] == 'Q') printf("%d\n", HLD_query(a, b));
else if(op[0] == 'C') HLD_update(a, b);
else HLD_negate(a, b);
}
}
return 0;
}