bzoj3730: 震波

題目傳送門

分析

題目大意：待修改詢問樹上距離某個點不超過K的所有點的點權和。
這種類似不超過多少距離和，樹上所有路徑之類的問題一般都是點分治，因為有修改所以是動態的。
對於每個分治重心開一顆以深度為鍵值的動態開點線段樹維護點權和。
每次在分治樹上跳即可。
注意因為需要刪除子樹影響，所以每個節點要額外維護一顆進入節點的子樹的線段樹。
複雜度 $O(n$

程式碼

#include<bits/stdc++.h>
const int N = 1e5 + 10;
int ri(
 
) {
    char c = getchar(); int x = 0, f = 1; for(;c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
    for(;c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) - '0' + c; return x * f;
}
int tp, n, sums, r1[N], r2[N], v[N];
int G, mn, pr[N], to[N <<
 
 1], nx[N << 1], sz[N], de[N], fa[N];
bool vis[N];
void add(int u, int v) {to[++tp] = v; nx[tp] = pr[u]; pr[u] = tp;}
void adds(int u, int v) {add(u, v); add(v, u);}
struct {
    int ls[N * 100], rs[N * 100], sum[N * 100], sz, val;
    void Modify(int &p, int L, int R, int x) {
        if(!p) p = ++sz; sum[p] += val;
        if(L == R) return ; int m = L + R >> 1;
        x <= m ? Modify(ls[p], L, m, x) : Modify(rs[p], m + 1, R, x);
    }
    int Query(int p, int L, int R, int k) {
        if(!p || k < L || k > R) return 0;
        if(R == k) return sum[p]; int m = L + R >> 1;
        return k <= m ? Query(ls[p], L, m, k) : sum[ls[p]] + Query(rs[p], m + 1, R, k);
    }
}seg;
struct {
    int bin[21], de[N], mn[21][N << 1], in[N], Lg[N << 1], tot;
    void Dfs(int u, int fa) {
        de[u] = de[fa] + 1; mn[0][++tot] = u; in[u] = tot;
        for(int i = pr[u]; i; i = nx[i])
            if(to[i] != fa) 
                Dfs(to[i], u), 
                mn[0][++tot] = u;
    }
    int Be(int x, int y) {return de[x] < de[y] ? x : y;}
    void Work() {
        Dfs(1, 0);
        bin[0] = 1; for(int i = 1;i <= 20; ++i) bin[i] = bin[i - 1] << 1;
        Lg[0] = -1; for(int i = 1;i <= tot; ++i) Lg[i] = Lg[i >> 1] + 1;
        for(int i = 1;bin[i] <= tot; ++i)
            for(int j = 1;j + bin[i] - 1 <= tot; ++j)
                mn[i][j] = Be(mn[i - 1][j], mn[i - 1][j + bin[i - 1]]);
    }
    int Lca(int u, int v) {
        u = in[u]; v = in[v];
        if(u > v) std::swap(u, v);
        int t = Lg[v - u + 1];
        return Be(mn[t][u], mn[t][v - bin[t] + 1]);
    }
    int Dist(int u, int v) {
        return de[u] + de[v] - (de[Lca(u, v)] << 1);
    }
}lca;
void Get(int u, int fa, int &R) {
    de[u] = de[fa] + 1; seg.val = v[u]; 
    seg.Modify(R, 0, n - 1, de[u]);
    for(int i = pr[u]; i; i = nx[i])
        if(to[i] != fa && !vis[to[i]])
            Get(to[i], u, R);
}
void Rt(int u, int fa) {
    sz[u] = 1; int tp = 0;
    for(int i = pr[u]; i; i = nx[i])
        if(to[i] != fa && !vis[to[i]]) {
            Rt(to[i], u);
            sz[u] += sz[to[i]];
            tp = std::max(tp, sz[to[i]]);
        }
    tp = std::max(tp, sums - sz[u]);
    if(mn > tp) mn = tp, G = u;
}
void Divide(int u, int in, int ps) {
    Get(u, 0, r1[u]); 
    if(in) Get(in, 0, r2[u]); 
    vis[u] = true;
    for(int i = pr[u]; i; i = nx[i])
        if(!vis[to[i]]) {
            sums = sz[to[i]] > sz[u] ? ps - sz[u] : sz[to[i]];
            G = 0; mn = 1e9; Rt(to[i], 0); fa[G] = u;
            Divide(G, to[i], sums);
        }
}
int Query(int u, int k) {
    int Ans = seg.Query(r1[u], 0, n - 1, k);
    for(int x = u; fa[x]; x = fa[x]) {
        int Dis = lca.Dist(u, fa[x]);
        Ans += seg.Query(r1[fa[x]], 0, n - 1, k - Dis);
        Ans -= seg.Query(r2[x], 0, n - 1, k - Dis - 1);
    }
    return Ans;
}
void Modify(int u, int val) {
    seg.val = val - v[u];
    seg.Modify(r1[u], 0, n - 1, 0);
    for(int x = u;x; x = fa[x]) {
        int Dis = lca.Dist(u, fa[x]);
        seg.Modify(r1[fa[x]], 0, n - 1, Dis);
        seg.Modify(r2[x], 0, n - 1, Dis - 1);
    }
    v[u] = val;
}
int main() {
    de[0] = -1; n = ri(); int m = ri();
    for(int i = 1;i <= n; ++i) 
        v[i] = ri();
    for(int i = 1;i < n; ++i) 
        adds(ri(), ri());
    lca.Work();
    G = 0; mn = 1e9; sums = n; Rt(1, 0); 
    Divide(G, 0, sums);
    for(int La = 0;m--; ) {
        int op = ri(), x = ri() ^ La, y = ri() ^ La;
        if(!op) printf("%d\n", La = Query(x, y));
        else Modify(x, y);
    }
    return 0;
}