fzu 2082(树链剖分)

题意：有一棵树，有n个节点，然后给出n-1条边的权值，然后有两种操作，1 a b 是要求把节点a到节点b路径上所有点的权值和输出，2 a b要求把边a的权值置为b。
题解：边权化点权，把每条边的权值都化为深度更大的点的点权，然后查询操作，当u和v的top成为一个点，再次查询是son[u]到v的权值和，因为u到v会多加一条边的权值。

#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;const int N = 50005;struct Edge {    int u, v, w, nxt;    Edge() {}    Edge(int a, int b, int c, int d): u(a), v(b), w(c), nxt(d) {}}e[N << 1];int top[N], son[N], fa[N], size[N], dep[N], id[N];int n, m, head[N], cnt, tot;ll tree[N << 2];void AddEdge(int u, int v, int w) {    e[cnt] = Edge(u, v, w, head[u]);    head[u] = cnt++;    e[cnt] = Edge(v, u, w, head[v]);    head[v] = cnt++;}void dfs2(int u, int tp) {    top[u] = tp;    id[u] = ++tot;    if (son[u]) dfs2(son[u], tp);    for (int i = head[u]; i + 1; i = e[i].nxt) {        int v = e[i].v;        if (v == fa[u] || v == son[u]) continue;        dfs2(v, v);    }}void dfs1(int u, int f, int depth) {    size[u] = 1, dep[u] = depth, son[u] = 0, fa[u] = f;    for (int i = head[u]; i + 1; i = e[i].nxt) {        int v = e[i].v;        if (v == f) continue;        dfs1(v, u, depth + 1);        size[u] += size[v];        if (size[v] > size[son[u]]) son[u] = v;    }}void pushup(int k) {    tree[k] = tree[k * 2] + tree[k * 2 + 1];}void modify(int k, int left, int right, int pos, ll v) {    if (left == right) {        tree[k] = v;        return;    }    int mid = (left + right) / 2;    if (pos <= mid)        modify(k * 2, left, mid, pos, v);    else        modify(k * 2 + 1, mid + 1, right, pos, v);    pushup(k);}ll query(int k, int left, int right, int l, int r) {    if (l <= left && right <= r)        return tree[k];    int mid = (left + right) / 2;    ll res = 0;    if (l <= mid)        res += query(k * 2, left, mid, l, r);    if (r > mid)        res += query(k * 2 + 1, mid + 1, right, l, r);    return res;}void solve(int u, int v) {    int tp1 = top[u], tp2 = top[v];    ll res = 0;    while (tp1 != tp2) {        if (dep[tp1] < dep[tp2]) {            swap(tp1, tp2);            swap(u, v);        }        res += query(1, 1, tot, id[tp1], id[u]);        u = fa[tp1];        tp1 = top[u];    }    if (u != v) {        if (dep[u] > dep[v]) swap(u, v);        res += query(1, 1, tot, id[son[u]], id[v]);    }    printf("%lld\n", res);}int main() {    while (scanf("%d%d", &n, &m) == 2) {        memset(head, -1, sizeof(head));        cnt = tot = 0;        int u, v;        ll w;        for (int i = 1; i <= n - 1; i++) {            scanf("%d%d%lld", &u, &v, &w);            AddEdge(u, v, w);        }        dfs1(1, 0, 1);        dfs2(1, 1);        memset(tree, 0, sizeof(tree));        for (int i = 0; i < cnt; i += 2) {            u = e[i].u, v = e[i].v;            if (dep[u] < dep[v])                swap(u, v);            modify(1, 1, tot, id[u], e[i].w);        }        int op, a;        ll b;        while (m--) {            scanf("%d%d%lld", &op, &a, &b);            if (op == 1)                solve(a, b);            else {                u = e[a - 1 << 1].u;                v = e[a - 1 << 1].v;                if (dep[u] < dep[v]) swap(u, v);                modify(1, 1, tot, id[u], b);            }        }    }    return 0;}