CF1111E Tree

过年了，洛咕还没爬这次的题，先放个CF的链接吧。

对于每个询问点\(x\)，设它的祖先即不能和它放在同一个集合中的点的个数为\(f[x]\)，设\(dp[i][j]\)表示前\(i\)个询问点放在\(j\)个非空集合中的方案数，注意这里“前\(i\)个”的意义，这表示会对第\(i\)个点造成影响的点都已被考虑过了，转移就是\(dp[i][j] = dp[i - 1][j] * (i - f[j]) + dp[i -1][j - 1]\)。

下面的问题就是怎么处理出\(f\)数组和找出DP的顺序。发现\(f\)数组可以直接树剖：先在线段树上把所有询问点更新一遍，然后再查询每个点到当前根的路径上询问点的个数，\(f[x]\)就是线段树上查询的值\(- 1\)（不算自己）。处理出\(f\)数组之后，发现祖先的\(f\)值一定比子孙的\(f\)值小，那么直接对\(f\)数组排一边序就可以DP了。

//written by newbiechd
#include <cstdio>
#include <cctype>
#include <vector>
#include <algorithm>
#define R register
#define I inline
#define B 1000000
#define L long long
using namespace std;
const int N = 100003, yyb = 1e9 + 7;
char buf[B], *p1, *p2;
I char gc() { return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, B, stdin), p1 == p2) ? EOF : *p1++; }
I int rd() {
    R int f = 0;
    R char c = gc();
    while (c < 48 || c > 57)
        c = gc();
    while (c > 47 && c < 58)
        f = f * 10 + (c ^ 48), c = gc();
    return f;
}
int a[N], s[N], fa[N], dep[N], siz[N], son[N], dfn[N], top[N], f[N], v[N << 2], n, tim;
L dp[N], ans;
vector <int> g[N];
void dfs1(int x, int f) {
    fa[x] = f, dep[x] = dep[f] + 1, siz[x] = 1;
    for (R int i = 0, y, m = 0; i < s[x]; ++i)
        if ((y = g[x][i]) ^ f) {
            dfs1(y, x), siz[x] += siz[y];
            if (siz[y] > m)
                m = siz[y], son[x] = y;
        }
}
void dfs2(int x, int t) {
    dfn[x] = ++tim, top[x] = t;
    if (son[x])
        dfs2(son[x], t);
    for (R int i = 0, y; i < s[x]; ++i)
        if ((y = g[x][i]) ^ fa[x] && y ^ son[x])
            dfs2(y, y);
}
void modify(int k, int l, int r, int x, int y) {
    v[k] += y;
    if (l == r)
        return ;
    R int p = k << 1, q = p | 1, m = l + r >> 1;
    if (x <= m)
        modify(p, l, m, x, y);
    else
        modify(q, m + 1, r, x, y);
}
int tquery(int k, int l, int r, int x, int y) {
    if (x <= l && r <= y)
        return v[k];
    R int p = k << 1, q = p | 1, m = l + r >> 1, o = 0;
    if (x <= m)
        o += tquery(p, l, m, x, y);
    if (m < y)
        o += tquery(q, m + 1, r, x, y);
    return o;
}
int query(int x, int y) {
    R int o = 0;
    while (top[x] ^ top[y]) {
        if (dep[top[x]] < dep[top[y]])
            swap(x, y);
        o += tquery(1, 1, n, dfn[top[x]], dfn[x]), x = fa[top[x]];
    }
    if (dep[x] > dep[y])
        swap(x, y);
    return o + tquery(1, 1, n, dfn[x], dfn[y]);
}
int main() {
    R int Q, k, m, rt, i, j, x, y, flag;
    n = rd(), Q = rd();
    for (i = 1; i < n; ++i)
        x = rd(), y = rd(), g[x].push_back(y), g[y].push_back(x);
    for (i = 1; i <= n; ++i)
        s[i] = g[i].size();
    dfs1(1, 0), dfs2(1, 1);
    while (Q--) {
        k = rd(), m = rd(), rt = rd(), ans = 0, flag = 0;
        for (i = 1; i <= k; ++i)
            a[i] = rd(), modify(1, 1, n, dfn[a[i]], 1);
        for (i = 1; i <= k; ++i) {
            f[i] = query(a[i], rt) - 1;
            if (f[i] >= m)
                flag = 1;
        }
        for (i = 1; i <= k; ++i)
            modify(1, 1, n, dfn[a[i]], -1), dp[i]  = 0;
        if (flag) {
            printf("0\n");
            continue;
        }
        sort(f + 1, f + k + 1), dp[0] = 1;
        for (i = 1; i <= k; ++i)
            for (j = min(i, m); ~j; --j) {
                if (j <= f[i])
                    dp[j] = 0;
                dp[j] = (dp[j] * (j - f[i]) + dp[j - 1]) % yyb;
            }
        for (j = 1; j <= k; ++j)
            ans = (ans + dp[j]) % yyb;
        printf("%I64d\n", ans);
    }
    return 0;
}

CF1111E Tree 树链剖分，DP

原文：https://www.cnblogs.com/cj-chd/p/10352143.html