BZOJ 3531 SDOI2014 旅行(travel) 树链剖分模板题

题目大意

给你一个有 N 个城市，构成一棵树。每个点有两个属性，一个是信仰(种类)，一个是评价(权值), 一条从城市 u 到城市 v 的路径上，只有与城市 u 信仰(种类)相同的城市才视为有效城市。
现在有 Q 个询问，每个询问读入三个数 Ord,u,v ，要求如下：
Ord=CC : 把城市 u 的点的信仰(种类)改为 v
Ord=CW : 把城市 u 的评价(权值)改为 v
Ord=QS : 询问从城市 u 到城市 v ，所有有效城市评价(权值)的和
Ord=QM : 询问从城市 u 到城市 v ，所有有效城市评价(权值)的和最大值

N<=100000 Q<=100000
宗教数小于100000 评价值小于10000

解题思路

用树链剖分为先预处理原树，再用线段树维链剖的最大值和权值和。每次找城市 u 和城市 v 的 Lca ,查找 u 到 Lca 和 v 到 Lca 的信息，再合并。

程序

//SDOI2014 旅行(travel) YxuanwKeith
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int MAXN = 2e5 + 5, MAXL = 17;

struct Tree { int Max, Sum, l, r;} Tr[MAXN * MAXL];

int N, M, tot, Val[MAXN], Bel[MAXN], Last[MAXN], Go[MAXN], Next[MAXN];
int Num, Cnt, Root[MAXN], Size[MAXN], Ord[MAXN], Top[MAXN], Mson[MAXN], Deep[MAXN], Fa[MAXN][MAXL + 1];

void Link(int u, int v) {
    Next[++ tot] = Last[u], Last[u] = tot, Go[tot] = v;
}

void Basis(int Now, int fa) {
    Size[Now] = 1, Fa[Now][0] = fa, Deep[Now] = Deep[fa] + 1;
    for (int p = Last[Now]; p; p = Next[p]) {
        int v = Go[p];
        if (v == fa) continue;
        Basis(v, Now);
        Size[Now] += Size[v];
        if (Size[Mson[Now]] < Size[v]) Mson[Now] = v;
    }

}

void GetFa() {
    for (int i = 1; i <= MAXL; i ++) 
        for (int j = 1; j <= N; j ++) 
            Fa[j][i] = Fa[Fa[j][i - 1]][i - 1];
}

void Promote(int Now, int top) {
    if (!Now) return;
    Ord[Now] = ++ Num,  Top[Now] = top;
    Promote(Mson[Now], top);
    for (int p = Last[Now]; p; p = Next[p]) {
        int v = Go[p];
        if (v == Fa[Now][0] || v == Mson[Now]) continue;
        Promote(v, v);
    }
}

int Lca(int x, int y) {
    if (Deep[x] < Deep[y]) swap(x, y);
    for (int i = MAXL; i + 1; i --)
        if (Deep[Fa[x][i]] >= Deep[y]) x = Fa[x][i];    
    if (x == y) return x;
    for (int i = MAXL; i + 1; i --)
        if (Fa[x][i] != Fa[y][i]) x = Fa[x][i], y = Fa[y][i];
    return Fa[x][0];
}

void Update(int Now) {
    int l = Tr[Now].l, r = Tr[Now].r;
    Tr[Now].Max = max(Tr[l].Max, Tr[r].Max);
    Tr[Now].Sum = Tr[l].Sum + Tr[r].Sum;
}   

void Modify(int &Now, int l, int r, int Num, int Val) {
    if (!Now) Now = ++ Cnt;
    if (l == r) {
        Tr[Now].Max = Tr[Now].Sum = Val;
        return;
    }
    int Mid = (l + r) >> 1;
    if (Num <= Mid) Modify(Tr[Now].l, l, Mid, Num, Val); else Modify(Tr[Now].r, Mid + 1, r, Num, Val);
    Update(Now);
}

int QueryS(int Now, int l, int r, int lx, int rx) {
    if (!Now) return 0;
    if (l == lx && r == rx) return Tr[Now].Sum;
    int Mid = (l + r) >> 1, L = Tr[Now].l, R = Tr[Now].r;
    if (rx <= Mid) return QueryS(L, l, Mid, lx, rx); else
    if (lx > Mid) return QueryS(R, Mid + 1, r, lx, rx); else
        return QueryS(L, l, Mid, lx, Mid) + QueryS(R, Mid + 1, r, Mid + 1, rx);
}

int QueryM(int Now, int l, int r, int lx, int rx) {
    if (!Now) return 0;
    if (l == lx && r == rx) return Tr[Now].Max;
    int Mid = (l + r) >> 1, L = Tr[Now].l, R = Tr[Now].r;
    if (rx <= Mid) return QueryM(L, l, Mid, lx, rx); else
    if (lx > Mid) return QueryM(R, Mid + 1, r, lx, rx); else
        return max(QueryM(L, l, Mid, lx, Mid), QueryM(R, Mid + 1, r, Mid + 1, rx));
}

void TreeChain() {
    Basis(1, 0), Promote(1, 1);
    GetFa();
    for (int i = 1; i <= N; i ++) Modify(Root[Bel[i]], 1, N, Ord[i], Val[i]);
}

int SolveSum(int ord, int u, int v) {
    int Sum = 0;
    for (; Top[u] != Top[v]; u = Fa[Top[u]][0])
        Sum += QueryS(Root[ord], 1, N, Ord[Top[u]], Ord[u]);
    Sum += QueryS(Root[ord], 1, N, Ord[v], Ord[u]);
    return Sum;
}

int SolveMax(int ord, int u, int v) {
    int Max = 0;
    for (; Top[u] != Top[v] ; u = Fa[Top[u]][0]) 
        Max = max(Max, QueryM(Root[ord], 1, N, Ord[Top[u]], Ord[u]));
    Max = max(Max, QueryM(Root[ord], 1, N, Ord[v], Ord[u]));
    return Max;
}

void GetSum(int u, int v) {
    int lca = Lca(u, v);
    int Ans = SolveSum(Bel[u], u, lca) + SolveSum(Bel[u], v, lca);
    if (Bel[u] == Bel[lca]) Ans -= Val[lca];
    printf("%d\n", Ans);
}

void GetMax(int u, int v) {
    int lca = Lca(u, v);
    printf("%d\n", max(SolveMax(Bel[u], u, lca), SolveMax(Bel[u], v, lca)));
}

int main() {
    freopen("travel.in", "r", stdin), freopen("travel.out", "w", stdout);

    scanf("%d%d", &N, &M);
    for (int i = 1; i <= N; i ++) scanf("%d%d", &Val[i], &Bel[i]);
    for (int i = 1; i < N; i ++) {
        int u, v;
        scanf("%d%d", &u, &v);
        Link(u, v), Link(v, u);
    }
    TreeChain();
    for (int i = 1; i <= M; i ++) {
        char C[2]; int u, v;
        scanf("%s%d%d", C, &u, &v);
        if (C[1] == 'W') Val[u] = v, Modify(Root[Bel[u]], 1, N, Ord[u], v);
        if (C[0] == 'C' && C[1] == 'C') {
            Modify(Root[Bel[u]], 1, N, Ord[u], 0);
            Bel[u] = v;
            Modify(Root[Bel[u]], 1, N, Ord[u], Val[u]);
        }
        if (C[1] == 'M') GetMax(u, v);
        if (C[0] == 'Q' && C[1] == 'S') GetSum(u, v);
    }
}