BZOJ1036: [ZJOI2008]树的统计Count (树链剖分)

传送门
这是一道树链剖分的基础题目，只需要用线段树来维护重链上的节点信息，轻链一条一条爬就行了，在查询的时候可以先求出LCA然后查询两个点到LCA的信息然后合并就行了。
代码：

/**************************************************************    Problem: 1036    User: geng4512    Language: C++    Result: Accepted    Time:3064 ms    Memory:5188 kb****************************************************************/#include<cstdio>#define MAXN 30005struct node {int v; node *nxt;} Edge[MAXN<<1], *Adj[MAXN], *M = Edge;inline void Add(int u, int v) { ++ M; M->v = v; M->nxt = Adj[u]; Adj[u] = M; }inline int Max(int a, int b) {return a > b ? a : b;}struct Seg{int l, r, mx, sum;} t[MAXN<<2];char c, f;inline void GET(int &n) {    f = 1; n = 0; do{c = getchar(); if(c == '-') f = -1; }while('0' > c || c > '9');    while('0' <= c && c <= '9') {n=n*10+c-'0';c=getchar();} n *= f;}int n, fa[MAXN], dep[MAXN], dcnt, pos[MAXN];int sz[MAXN], htp[MAXN], hsn[MAXN], a[MAXN];int px[MAXN], q, rev[MAXN];/************ 线段树相关 ******************/void Build(int i, int l, int r) {    t[i].l = l; t[i].r = r;    if(l == r) {        px[l] = i; t[i].sum = t[i].mx = a[rev[l]]; return;    }    int mid = (l+r)>>1;    Build(i<<1, l, mid); Build(i<<1|1, mid+1, r);    t[i].mx = Max(t[i<<1].mx, t[i<<1|1].mx);    t[i].sum = t[i<<1].sum + t[i<<1|1].sum;}void Ins(int x, int v) {    int p = px[x]; t[p].mx = t[p].sum = v; p >>= 1;    while(p) {        t[p].mx = Max(t[p<<1].mx, t[p<<1|1].mx);        t[p].sum = t[p<<1].sum + t[p<<1|1].sum;        p >>= 1;    }}int Qmx(int i, int L, int R) {    if(t[i].r < L || t[i].l > R) return -0x3f3f3f3f;    if(L <= t[i].l && t[i].r <= R) return t[i].mx;    return Max(Qmx(i<<1, L, R), Qmx(i<<1|1, L, R));}int Qsm(int i, int L, int R) {    if(t[i].r < L || t[i].l > R) return 0;    if(L <= t[i].l && t[i].r <= R) return t[i].sum;    return Qsm(i<<1, L, R) + Qsm(i<<1|1, L, R);}/****************************************/int tmp;inline void Swap(int &a, int &b) {tmp = a; a = b; b = tmp;}inline int LCA(int a, int b) {    while(htp[a] != htp[b]) {        if(dep[htp[a]] < dep[htp[b]]) Swap(a, b);        a = fa[htp[a]];    }    return dep[a] < dep[b] ? a : b;}inline int sum(int a, int b) {    int sum = 0;    while(htp[a] != htp[b]) {    //查重链，爬轻链        sum += Qsm(1, pos[htp[a]], pos[a]);        a = fa[htp[a]];    }    return sum += Qsm(1, pos[b], pos[a]);}inline int Mx(int a, int b) {    int Mx = -0x3f3f3f3f;    while(htp[a] != htp[b]) {        Mx = Max(Mx, Qmx(1, pos[htp[a]], pos[a]));        a = fa[htp[a]];    }    Mx = Max(Qmx(1, pos[b], pos[a]), Mx);    return Mx;}/*************** 轻重链剖分 ***************/void dfs1(int u) {    sz[u] = 1;    for(node *p = Adj[u]; p; p = p->nxt) {        if(sz[p->v]) continue;        fa[p->v] = u; dep[p->v] = dep[u]+1;        dfs1(p->v);        if(sz[hsn[u]] < sz[p->v]) hsn[u] = p->v;        sz[u] += sz[p->v];    }}void dfs2(int u, int tp) {    pos[u] = ++ dcnt; htp[u] = tp; rev[dcnt] = u; //pos记录了u的dfs序。    if(!hsn[u]) return;    dfs2(hsn[u], tp);    for(node *p = Adj[u]; p; p = p->nxt)        if(htp[p->v]) continue;        else if(p->v != hsn[u]) dfs2(p->v, p->v);}/******************************************/int main() {    GET(n); int u, v;    for(int i = 1; i < n; ++ i) {        GET(u); GET(v);        Add(u, v); Add(v, u);    }    for(int i = 1; i <= n; ++ i) GET(a[i]);    dfs1(1); dfs2(1, 1);    Build(1, 1, n);    GET(q); char s[6];    for(int i = 1; i <= q; ++ i) {        scanf("%s", s); GET(u); GET(v);        if(s[0] == 'C') { a[u] = v; Ins(pos[u], v); }        else {            int t = LCA(u, v);            if(s[1] == 'M') printf("%d\n", Max(Mx(u, t), Mx(v, t)));            else printf("%d\n", sum(u, t) + sum(v, t) - a[t]);        }    }    return 0;}