原题传送门
最近在打树剖练手题啦
此题与原始树剖不太一样,本来我们维护的是点权,现在我们维护的是边权,怎么办?
其实只要让每一条边的两个顶点深度大的那个代表此边,这样一来,我们就可以通过维护“点权”来维护边权
其余的都是跟模板一样的
Code:
#include <bits/stdc++.h>
#define maxn 300010
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
struct Edge{
int to, next, len;
}edge[maxn << 1];
struct Seg{
int l, r, sum;
}seg[maxn << 2];
int head[maxn << 1], d[maxn], fa[maxn], size[maxn], son[maxn], top[maxn], id[maxn], w[maxn], wt[maxn], num, cnt, n;
struct Line{
int x, y, z;
}line[maxn];
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
void add_edge(int x, int y, int z){ edge[++num].to = y; edge[num].len = z; edge[num].next = head[x]; head[x] = num; }
void pushup(int rt){ seg[rt].sum = max(seg[ls].sum, seg[rs].sum); }
void build(int rt, int l, int r){
seg[rt].l = l, seg[rt].r = r;
if (l == r){
seg[rt].sum = wt[l]; return;
}
int mid = (l + r) >> 1;
build(ls, l, mid); build(rs, mid + 1, r);
pushup(rt);
}
void update(int rt, int x, int y){
if (seg[rt].l == seg[rt].r){
seg[rt].sum = y; return;
}
if (seg[ls].r >= x) update(ls, x, y); else update(rs, x, y);
pushup(rt);
}
int query(int rt, int l, int r){
if (seg[rt].l > r || seg[rt].r < l) return 0;
if (seg[rt].l >= l && seg[rt].r <= r) return seg[rt].sum;
return max(query(ls, l, r), query(rs, l, r));
}
int query_range(int u, int v){
int sum = 0;
while (top[u] != top[v]){
if (d[top[u]] < d[top[v]]) swap(u, v);
sum = max(sum, query(1, id[top[u]], id[u]));
u = fa[top[u]];
}
if (d[u] < d[v]) swap(u, v);
return max(sum, query(1, id[v] + 1, id[u]));
}
void dfs1(int u){
size[u] = 1, son[u] = -1;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != fa[u]){
fa[v] = u, d[v] = d[u] + 1, w[v] = edge[i].len;
dfs1(v);
size[u] += size[v];
if (son[u] == -1) son[u] = v; else
if (size[son[u]] < size[v]) son[u] = v;
}
}
}
void dfs2(int u, int x){
id[u] = ++cnt, top[u] = x, wt[cnt] = w[u];
if (son[u] == -1) return;
dfs2(son[u], x);
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != fa[u] && v != son[u]) dfs2(v, v);
}
}
int main(){
n = read();
for (int i = 1; i < n; ++i){
int x = read(), y = read(), z = read();
add_edge(x, y, z); add_edge(y, x, z);
line[i].x = x; line[i].y = y; line[i].z = z;
}
dfs1(1);
dfs2(1, 1);
build(1, 1, n);
while (1){
char c = getchar();
for (; c != 'Q' && c != 'C' && c != 'D'; c = getchar());
if (c == 'D') break;
int x = read(), y = read();
if (c == 'Q') printf("%d\n", query_range(x, y)); else{
line[x].z = y;
int u = line[x].x, v = line[x].y;
if (fa[u] == v) swap(u, v);
update(1, id[v], y);
}
}
return 0;
}