CF 1260F colored tree

令 \(p=\prod (r_i-l_i+1),len_x=r_x-l_x+1\)。

考虑枚举颜色和点对，写出答案的式子：

\[\sum _c\sum_x[l_x\leq c\leq r_x]\sum _ y [l_y\leq c\leq r_y] dis(x,y)\times \frac {p} {len_x\times len_y} \]

将 \(dis(x,y)\) 用深度代替，式子变为：

\[\sum _c \sum _x [l_x\leq c\leq r_x] \sum _y[l_y\leq c\leq r_y] \frac {p\times (dep[x]+dep[y]-2\times dep[lca(x,y)])} {len_x\times len_y} \]

整理式子，得到：

\[\sum _ c \sum _x [l_x\leq c\leq r_y] \frac p{len_x}\sum _y [l_y\leq c\leq r_y]\frac {dep[x]}{len_y}+\frac {dep[y]}{len_y}-2\frac {dep[lca(x,y)]}{len_y} \]

从小到大枚举 \(c\)，将点逐个加入或删除，维护颜色为 \(c\) 时的答案。

发现加入一个点 \(z\) 时只需要维护 \(\sum \frac 1 {len_x}\)，\(\sum \frac {dep[x]}{len_x}\) 和 \(\sum \frac {dep[lca(x,z)]}{len_x}\)。

前两个很好维护，最后一个可以在加入点 \(x\) 时把 \(x\to 1\) 的路径上加 \(\frac 1{len_x}\)。查询时查询 \(z\to 1\) 的路径和即可。可以使用树剖或 \(LCT\) 维护。和 P4211 LNOI2014 LCA 的方法类似。

复杂度：\(O(n\log ^2n)\) 或 \(O(n\log n)\)。

代码

#include <bits/stdc++.h>

const int N = 1e5 + 10, Mod = 1e9 + 7;
#define Mod 1000000007
#define pii std::pair<int, int>
#define mkp(a, b) std::make_pair(a, b)
typedef struct { int p, x, f; } node;

inline int read()
{
    int x = 0;
    char ch = getchar();
    while (!isdigit(ch)) ch = getchar();
    while ( isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return x;
}
inline int Pow(int a, int b, int p = Mod)
{
    int res = 1;
    for (; b; b >>= 1, a = 1LL * a * a % p)
        if (b & 1) res = 1LL * res * a % p;
    return res;
}

std::vector<pii> vec[N];

struct Node
{
    int len, inv;
    inline Node(int dat = 0)
    {
        len = dat, inv = Pow(dat, Mod - 2);
    }
} a[N];

struct SegmentTree
{
    #define mid ((l + r) >> 1)
    int tag[N << 2], sum[N << 2];

    inline void upd(int k) 
    { 
        sum[k] = (sum[k << 1] + sum[k << 1 | 1]) % Mod; 
    }

    inline void down(int k, int l, int r)
    {
        if (!tag[k]) return;
        (sum[k << 1 | 1] += 1LL * tag[k] * (r - mid) % Mod) %= Mod;
        (sum[k << 1] += 1LL * tag[k] * (mid - l + 1) % Mod) %= Mod;
        (tag[k << 1] += tag[k]) %= Mod, (tag[k << 1 | 1] += tag[k]) %= Mod;
        tag[k] = 0;
    }

    inline void Add(int k, int l, int r, int x, int y, int dat)
    {
        if (y < l || r < x) return;
        if (x <= l && r <= y)
            return (sum[k] += 1LL * (r - l + 1) * dat % Mod) %= Mod, (tag[k] += dat) %= Mod, void();
        down(k, l, r), Add(k << 1, l, mid, x, y, dat), Add(k << 1 | 1, mid + 1, r, x, y, dat), upd(k);
    }
    inline int Ask(int k, int l, int r, int x, int y)
    {
        if (y < l || r < x) return 0;
        if (x <= l && r <= y) return sum[k];
        down(k, l, r);
        return (Ask(k << 1, l, mid, x, y) + Ask(k << 1 | 1, mid + 1, r, x, y)) % Mod;
    }
} T;

int inv[N];
int tot = 1, fir[N], nex[N << 1], got[N << 1];

inline void AddEdge(int x, int y)
{
    nex[++tot] = fir[x], fir[x] = tot, got[tot] = y;
}

int siz[N], dep[N], par[N], son[N];

inline void dfs1(int x, int fa)
{
    siz[x] = 1, dep[x] = dep[par[x] = fa] + 1;
    for (int i = fir[x]; i; i = nex[i])
        if (got[i] != fa)
        {
            dfs1(got[i], x), siz[x] += siz[got[i]];
            if (siz[got[i]] >= siz[son[x]]) son[x] = got[i];
        }
}

int top[N], dfn[N], idx;

inline void dfs2(int x, int fa)
{
    top[x] = (x == son[fa] ? top[fa] : x), dfn[x] = ++idx;
    if (son[x]) dfs2(son[x], x);
    for (int i = fir[x]; i; i = nex[i])
        if (got[i] != fa && got[i] != son[x]) dfs2(got[i], x);
}
inline void Add(int x, int n, int dat)
{
    while (x)
        T.Add(1, 1, n, dfn[top[x]], dfn[x], dat), x = par[top[x]];
}
inline int Ask(int x, int n)
{
    int ans = 0;
    while (x)
        (ans += T.Ask(1, 1, n, dfn[top[x]], dfn[x])) %= Mod, x = par[top[x]];
    return ans;
}
int main()
{
    int n = read(), ans = 0, C = 0, p = 1;
    for (int i = 1; i <= n; i++)
    {
        int l = read(), r = read();
        vec[l].push_back(mkp(i, 1)), vec[r + 1].push_back(mkp(i, -1));
        inv[i] = Pow(r - l + 1, Mod - 2, Mod), C = std::max(C, r), p = 1LL * p * (r - l + 1) % Mod;
    }
    for (int i = 1; i <= n - 1; i++)
    {
        int x = read(), y = read();
        AddEdge(x, y), AddEdge(y, x);
    }
    dfs1(1, 0), dfs2(1, 0);
    int sum1 = 0, sum2 = 0, s = 0;
    for (int i = 1; i <= C + 1; i++)
    {
        for (auto k : vec[i])
        {
            int x = k.first, len = inv[x];
            if (k.second == -1) len = Mod - len;
            (sum1 += len) %= Mod, (sum2 += 1LL * dep[x] * len % Mod) %= Mod, Add(x, n, len);
            (s += 1LL * p * len % Mod * (1LL * dep[x] * sum1 % Mod + sum2 - 2LL * Ask(x, n) % Mod + Mod) % Mod) %= Mod;
        }
        (ans += s) %= Mod;
    }
    printf("%d\n", ans);
    return 0;
}