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\(dp[i][j] = dp[i - 1][j] * (j - f[j]) + dp[i -1][j - 1]\)。

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\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;

}

巴特西

CF1111E Tree 树链剖分，DP

CF1111E Tree

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