【BZOJ5293】【BJOI2018】求和（LCA，树上差分）

Description

master 对树上的求和非常感兴趣。他生成了一棵有根树，并且希望多次询问这棵树上一段路径上所有节点深度的k 次方和，而且每次的k 可能是不同的。此处节点深度的定义是这个节点到根的路径上的边数。他把这个问题交给了pupil，但pupil 并不会这么复杂的操作，你能帮他解决吗？

Solution

树上差分傻逼题。

对于不同的 $k$ 分别处理，直接累一个到根的前缀和，然后 $sum[u]+sum[v]-sum[lca]-sum[fa[lca]]$ 即可。

Code

/************************************************
 * Au: Hany01
 * Date: Sep 5th, 2018
 * Prob: BZOJ5293 BJOI2018 求和
 * Email: [email protected] & [email protected]
 * Inst: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define ALL(a) a.begin(), a.end()
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    static int _, __; static char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 3e5 + 5, MOD = 998244353;

int n, beg[maxn], nex[maxn << 1], v[maxn << 1], e = 1, val[51][maxn], fa[19][maxn], dep[maxn];

inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }
inline int ad(int x, int y) { return (x += y) >= MOD ? x - MOD : x; }

void DFS(int u, int pa, int depth) {
    int tmp = depth; fa[0][u] = pa, dep[u] = depth + 1;
    For(i, 1, 50) val[i][u] = ad(val[i][pa], tmp), tmp = (LL)tmp * depth % MOD;
    For(i, 1, 18) fa[i][u] = fa[i - 1][fa[i - 1][u]];
    for (register int i = beg[u]; i; i = nex[i])
        if (v[i] != pa) DFS(v[i], u, depth + 1);
}

inline int LCA(int u, int v) {
    if (dep[u] < dep[v]) swap(u, v);
    Fordown(i, 18, 0) if (dep[fa[i][u]] >= dep[v]) u = fa[i][u];
    if (u == v) return u;
    Fordown(i, 18, 0) if (fa[i][u] ^ fa[i][v]) u = fa[i][u], v = fa[i][v];
    return fa[0][u];
}

inline int query(int u, int v, int k) {
    static int lca; lca = LCA(u, v);
    return ad(ad(ad(val[k][u], val[k][v]), MOD - val[k][lca]), MOD - val[k][fa[0][lca]]);
}

int main()
{
#ifdef hany01
    freopen("bzoj5293.in", "r", stdin);
    freopen("bzoj5293.out", "w", stdout);
#endif

    static int u, v, k;

    n = read();
    For(i, 2, n) u = read(), v = read(), add(u, v), add(v, u);
    DFS(1, 0, 0);
    for (static int m = read(); m --; )
        u = read(), v = read(), k = read(), printf("%d\n", query(u, v, k));

    return 0;
}

【BZOJ5293】【BJOI2018】求和（LCA，树上差分）

Description

Solution

Code

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