题目链接：传送门
分析一下他给的这个图就知道 一个祖先到它的儿子只有生成树的边可以走 而一个儿子走到祖先有两种方法 一个是先直接到1，再从1走到那个祖先 还有一个很容易想不到，就是继续往它的儿子走，再从它的某个儿子回到1 比如说我一开始就没想到，WA了半天才发现这个问题 所以还需要维护这样的一个最小值 没心情改了，先把没改的代码放这
#include <bits/stdc++.h> #define A 1000010 using namespace std; typedef long long ll; struct node {int next, to, w;}e[A]; int head[A], num; void add(int fr, int to, int w) {e[++num].next = head[fr]; e[num].to = to; e[num].w = w; head[fr] = num;} int siz[A], fa[A], w[A], dep[A], son[A], dfn[A], cnt, top[A], pre[A]; void prepare(int fr) { siz[fr] = 1; for (int i = head[fr]; i; i = e[i].next) { int ca = e[i].to; if (ca == fa[fr]) continue; fa[ca] = fr; w[ca] = e[i].w; dep[ca] = dep[fr] + 1; prepare(ca); siz[fr] += siz[ca]; if (siz[ca] > siz[son[fr]]) son[fr] = ca; } } void dfs(int fr, int tp) { dfn[fr] = ++cnt; pre[cnt] = fr; top[fr] = tp; if (son[fr]) dfs(son[fr], tp); for (int i = head[fr]; i; i = e[i].next) { int ca = e[i].to; if (ca == fa[fr] or ca == son[fr]) continue; dfs(ca, ca); } } struct Tree {int l, r, w;}tree[A]; void build(int k, int l, int r) { tree[k].l = l; tree[k].r = r; if (l == r) {tree[k].w = w[pre[l]]; return;} int m = (l + r) >> 1; build(k << 1, l, m); build(k << 1 | 1, m + 1, r); tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w; } void change(int k, int pos, int val) { if (tree[k].l == tree[k].r) {tree[k].w = val; return;} int m = (tree[k].l + tree[k].r) >> 1; if (pos <= m) change(k << 1, pos, val); else change(k << 1 | 1, pos, val); tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w; } int asktree(int k, int l, int r) { if (tree[k].l >= l and tree[k].r <= r) return tree[k].w; int m = (tree[k].l + tree[k].r) >> 1, ans = 0; if (l <= m) ans += asktree(k << 1, l, r); if (r > m) ans += asktree(k << 1 | 1, l, r); return ans; } int ask(int x, int y, int ans = 0) { while (top[x] != top[y]) { if (dep[top[x]] < dep[top[y]]) swap(x, y); ans += asktree(1, dfn[top[x]], dfn[x]); x = fa[top[x]]; } if (dep[x] > dep[y]) swap(x, y); ans += asktree(1, dfn[x] + 1, dfn[y]); return ans; } map<int, int> v[A]; int n, q, a[A], b[A], c[A], opt, x, y; int main(int argc, char const *argv[]) { cin >> n >> q; for (int i = 1; i < n; i++) { scanf("%d%d%d", &a[i], &b[i], &c[i]); add(a[i], b[i], c[i]); } prepare(1); dfs(1, 1); build(1, 1, n); for (int i = n; i < 2 * n - 1; i++) { scanf("%d%d%d", &a[i], &b[i], &c[i]); v[a[i]][b[i]] = c[i]; } while (q--) { scanf("%d%d%d", &opt, &x, &y); if (opt == 1) { if (x <= n - 1) change(1, dfn[b[x]], y); else v[a[x]][b[x]] = y; } else { if (x == y) {puts("0"); continue;} if (dfn[x] < dfn[y] and dfn[x] + siz[x] > dfn[y]) {printf("%d\n", ask(x, y)); continue;} printf("%d\n", v[x][1] + ask(1, y)); } } }