hysbz 2243(树链剖分+区间合并)

题意：有一棵树，n个节点，n-1条边，每个点都有一个权值，现在有两种操作，Q a b，询问a到b路径上所有点组成的序列一共有几段，比如11223就是三段，C a b c把节点a到b路径上所有经过的点权值设为c。
题解：线段树的区间合并问题，用线段树维护每个区间的左端点值和右端点值还有区间内有几段。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 100005;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], val2[N], size[N], son[N], top[N], id[N], dep[N], fa[N];int n, m, cnt, tot, head[N], tree[N << 2], flag[N << 2], lv[N << 2], rv[N << 2];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, son[u] = 0, dep[u] = depth, 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) {    lv[k] = lv[k * 2];      rv[k] = rv[k * 2 + 1];    tree[k] = tree[k * 2] + tree[k * 2 + 1];    if (rv[k * 2] == lv[k * 2 + 1])        tree[k]--;}void pushdown(int k) {    if (flag[k] >= 0) {        tree[k * 2] = tree[k * 2 + 1] = 1;        lv[k * 2] = lv[k * 2 + 1] = flag[k];        rv[k * 2] = rv[k * 2 + 1] = flag[k];        flag[k * 2] = flag[k * 2 + 1] = flag[k];        flag[k] = -1;    }}void build(int k, int left, int right) {    flag[k] = -1;    if (left == right) {        lv[k] = rv[k] = val2[left];        tree[k] = 1;        return;    }    int mid = (left + right) / 2;    build(k * 2, left, mid);    build(k * 2 + 1, mid + 1, right);    pushup(k);}void modify(int k, int left, int right, int l, int r, int v) {    if (l <= left && right <= r) {        lv[k] = rv[k] = flag[k] = v;        tree[k] = 1;        return;    }    pushdown(k);    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);}int query(int k, int left, int right, int l, int r) {    if (l <= left && right <= r)        return tree[k];    pushdown(k);    int mid = (left + right) / 2;    if (r <= mid)        return query(k * 2, left, mid, l, r);    if (l > mid)        return query(k * 2 + 1, mid + 1, right, l, r);    int res = query(k * 2, left, mid, l, mid) + query(k * 2 + 1, mid + 1, right, mid + 1, r);    if (rv[k * 2] == lv[k * 2 + 1])        res--;    return res;}int query2(int k, int left, int right, int pos) {    if (left == right)        return lv[k];    pushdown(k);    int mid = (left + right) / 2;    if (pos <= mid)        return query2(k * 2, left, mid, pos);    return query2(k * 2 + 1, mid + 1, right, pos);}void solve(int u, int v, int w, int flag1) {    int tp1 = top[u], tp2 = top[v];     int res = 0;    while (tp1 != tp2) {        if (dep[tp1] < dep[tp2]) {            swap(tp1, tp2);            swap(u, v);        }        if (!flag1) {            res += query(1, 1, tot, id[tp1], id[u]);            int temp1 = query2(1, 1, tot, id[fa[tp1]]);            int temp2 = query2(1, 1, tot, id[tp1]);             if (temp1 == temp2)                res--;        }        else modify(1, 1, tot, id[tp1], id[u], w);        u = fa[tp1];        tp1 = top[u];    }    if (dep[u] > dep[v]) swap(u, v);    if (!flag1)        res += query(1, 1, tot, id[u], id[v]);     else modify(1, 1, tot, id[u], id[v], w);    if (!flag1) printf("%d\n", res);}int main() {    while (scanf("%d%d", &n, &m) == 2) {        memset(head, -1, sizeof(head));        cnt = tot = 0;        for (int i = 1; i <= n; i++)            scanf("%d", &val[i]);        int u, v;        for (int i = 0; i < n - 1; i++) {            scanf("%d%d", &u, &v);            AddEdge(u, v);        }        dfs1(1, 0, 1);        dfs2(1, 1);        for (int i = 1; i <= n; i++)            val2[id[i]] = val[i];        build(1, 1, tot);        char op[5];        int a, b, c;        while (m--) {            scanf("%s%d%d", op, &a, &b);            if (op[0] == 'Q') solve(a, b, 0, 0);            else {                scanf("%d", &c);                solve(a, b, c, 1);            }        }    }    return 0;}