传送门:点击打开链接
题意:边更新,路径查询边权最大值
思路:第一道树链剖分题,其实树链剖分就相当于把树的边给分类了一样,分成了重边和轻边。
然后,有一个性质,重链的条数和轻边的条数都不会超过log(n),所以总的复杂也只有log(n)
也就是说,本身要维护一个路径上的某个东西,如果直接用线段树做,并不是很好做,
树链剖分就相当于把重链和轻边分开求,因为重链在DFS序下编号是可以连在一起的,所以重链是整条都被另一种数据结构求出了答案,轻边是单独的被求出答案,然后两部分答案合并起来~
听起来很复杂,但是实际操作起来却非常的神奇,代码也并没有那么繁琐,但是却十分十分的巧妙~
#include<bits/stdc++.h> #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].u = u; E[h_r].v = v; E[h_r].cost = cost; E[h_r].nxt = Head[u]; Head[u] = h_r++; } int MAX[MX << 2], A[MX]; void push_up(int rt) { MAX[rt] = max(MAX[rt << 1], MAX[rt << 1 | 1]); } void build(int l, int r, int rt) { if(l == r) { MAX[rt] = A[l]; return; } int m = (l + r) >> 1; build(lson); build(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; if(L <= m) ret = max(ret, query(L, R, lson)); if(R > m) ret = max(ret, query(L, R, rson)); return ret; } void update(int x, int d, int l, int r, int rt) { if(l == r) { MAX[rt] = d; return; } int m = (l + r) >> 1; if(x <= m) update(x, d, lson); else update(x, d, rson); push_up(rt); } 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); } } 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); } 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); } 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 HLD_update(a, b); } } return 0; }