CF1111E Tree
过年了，洛咕还没爬这次的题，先放个CF的链接吧。
补个LG传送门。
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] * (j - 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
");
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
", ans);
}
return 0;
}