[题解]LuoGu4114：Qtree1

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;
}