hdu 3966(树链剖分)

题意：有一棵树，上面n个节点，每个节点都有一个权值，有三种操作，I a b c，把从节点a到节点b路径上所有点(包括a和b)权值都加c，D a b c，把从节点a到节点b路径上所有点权值都减c，Q a，问节点a的权值。
题解：树链剖分模板题，线段树维护区间和，用到单点查询，区间更新。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 50005;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], top[N], size[N], fa[N], dep[N], son[N];int n, m, q, head[N], cnt, tree[N << 2], flag[N << 2], id[N], tot;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 pushup(int k) {    tree[k] = tree[k * 2] + tree[k * 2 + 1];}void pushdown(int k, int left, int right) {    if (flag[k] != 0) {        int mid = (left + right) / 2;        tree[k * 2] += flag[k] * (mid - left + 1);        flag[k * 2] += flag[k];        tree[k * 2 + 1] += flag[k] * (right - mid);        flag[k * 2 + 1] += flag[k];        flag[k] = 0;    }}void build(int k, int left, int right) {    tree[k] = flag[k] = 0;    if (left != right) {        int mid = (left + right) / 2;        build(k * 2, left, mid);        build(k * 2 + 1, mid + 1, right);    }}int query(int k, int left, int right, int pos) {    if (left == right)        return tree[k];    pushdown(k, left, right);    int mid = (left + right) / 2;    if (pos <= mid)        return query(k * 2, left, mid, pos);    return query(k * 2 + 1, mid + 1, right, pos);}void modify(int k, int left, int right, int l, int r, int v) {    if (l <= left && right <= r) {        tree[k] += (right - left + 1) * v;        flag[k] += v;        return;    }    pushdown(k, left, right);    int mid = (left + right) / 2;    if (l <= mid)        modify(k * 2, left, mid, l, r, v);    if (r > mid)        modify(k * 2 + 1, mid + 1, right, l, r, v);    pushup(k);}void dfs1(int u, int f, int depth) {    size[u] = 1; fa[u] = f; dep[u] = depth; son[u] = 0;    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[son[u]] < size[v]) son[u] = v;        }}void dfs2(int u, int tp) {    id[u] = ++tot;    top[u] = tp;    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 init() {    for (int i = 1; i <= n; i++)        scanf("%d", &val[i]);    memset(head, -1, sizeof(head));    cnt = tot = 0;    int u, v;    for (int i = 0; i < m; i++) {        scanf("%d%d", &u, &v);        AddEdge(u, v);    }    dfs1(1, 0, 1);    dfs2(1, 1);    build(1, 1, tot);    for (int i = 1; i <= n; i++)        modify(1, 1, tot, id[i], id[i], val[i]);}void solve(int u, int v, int a) {    int tp1 = top[u], tp2 = top[v];    int res = 0;    while (top[u] != top[v]) {        if (dep[tp1] < dep[tp2]) {            swap(tp1, tp2);            swap(u, v);        }        modify(1, 1, tot, id[tp1], id[u], a);        u = fa[tp1];        tp1 = top[u];    }    if (dep[u] > dep[v])        swap(u, v);    modify(1, 1, tot, id[u], id[v], a);}int main() {    while (scanf("%d%d%d", &n, &m, &q) == 3) {        init();        char op[5];        int a, b, c;        while (q--) {            scanf("%s", op);            if (op[0] == 'Q') {                scanf("%d", &a);                printf("%d\n", query(1, 1, tot, id[a]));            }            else if (op[0] == 'I') {                scanf("%d%d%d", &a, &b, &c);                solve(a, b, c);            }            else {                scanf("%d%d%d", &a, &b, &c);                solve(a, b, -c);            }        }    }    return 0;}