LightOj 1348(树链剖分)

题意：一棵树，有n个点，每个点都有一个权值，有两种操作，0 a b ，问从节点a到节点b路径上所有点权值和，1 a b，把节点a权值改为b。
题解：树链剖分水题。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 30005;struct Edge {    int u, v, nxt;    Edge() {};    Edge(int a, int b, int c): u(a), v(b), nxt(c) {}}e[N << 1];int val[N], size[N], dep[N], son[N], fa[N], top[N];int n, q, cnt, tot, head[N], tree[N << 2], id[N];void AddEdge(int u, int v) {    e[cnt] = Edge(u, v, head[u]);    head[u] = cnt++;    e[cnt] = Edge(v, u, 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, int 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);}int query(int k, int left, int right, int l, int r) {    if (l <= left && right <= r)        return tree[k];    int mid = (left + right) / 2, 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 init() {    memset(head, -1, sizeof(head));    cnt = tot = 0;    int u, v;    for (int i = 0; i < n - 1; i++) {        scanf("%d%d", &u, &v);        AddEdge(u + 1, v + 1);    }    dfs1(1, 0, 1);    dfs2(1, 1);    memset(tree, 0, sizeof(tree));    for (int i = 1; i <= n; i++)        modify(1, 1, tot, id[i], val[i]);}void solve(int u, int v) {    int tp1 = top[u], tp2 = top[v];     int 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 (dep[u] > dep[v]) swap(u, v);    res += query(1, 1, tot, id[u], id[v]);      printf("%d\n", res);}int main() {    int t, cas = 1;    scanf("%d", &t);    while (t--) {        scanf("%d", &n);        for (int i = 1; i <= n; i++)            scanf("%d", &val[i]);        init();        scanf("%d", &q);        int op, a, b;        printf("Case %d:\n", cas++);        while (q--) {            scanf("%d%d%d", &op, &a, &b);            if (op == 1)                modify(1, 1, tot, id[a + 1], b);            else                solve(a + 1, b + 1);        }    }    return 0;}