hdu5909 Tree Cutting

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=5909

【题解】

设$f_{x,i}$表示以$x$节点的子树中，权值为$i$的子树个数，其中$x$必选。

那么有dp方程：$f_{x,i} = sum_{y = son[x]} f_{x,i} + sum_{j oplus k = i}f_{x, j}f_{y, k}$

用FWT优化转移即可，复杂度$O(nmlogm)$。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int N = 1e3 + 5, H = 1024 + 5;
const int mod = 1e9+7;

int n, L, w[N], inv2;
int f[N][H];
int ans[H];
int head[N], nxt[N + N], to[N + N], tot = 0;
inline void add(int u, int v) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v;
}
inline void adde(int u, int v) {
    add(u, v), add(v, u);
}

inline int pwr(int a, int b) {
    int ret = 1;
    while(b) {
        if(b&1) ret = 1ll * ret * a % mod;
        a = 1ll * a * a % mod;
        b >>= 1;
    }
    return ret;
}

int s[H], t[H];
inline void FWT(int *a, int op) {
    if(op) {
        for (int len = 2; len <= L; len<<=1) {
            int m = len >> 1;
            for (int *p = a; p != a+L; p += len) {
                for (int k=0; k<m; ++k) {
                    int x = p[k], y = p[k+m];
                    p[k] = 1ll * (x+y) * inv2 % mod;
                    p[k+m] = 1ll * (x-y+mod) * inv2 % mod;
                }
            }
        }
    } else {
        for (int len = 2; len <= L; len<<=1) {
            int m = len >> 1;
            for (int *p = a; p != a+L; p += len) {
                for (int k=0; k<m; ++k) {
                    int x = p[k], y = p[k+m];
                    p[k] = (x+y) % mod;
                    p[k+m] = (x-y+mod) % mod;
                }
            }
        }
    }
}

inline void FWT_combine(int *A, int *B) {
    for (int i=0; i<L; ++i) s[i] = A[i], t[i] = B[i];
    FWT(s, 0); FWT(t, 0);
    for (int i=0; i<L; ++i) s[i] = 1ll * s[i] * t[i] % mod;
    FWT(s, 1);
    for (int i=0; i<L; ++i) (A[i] += s[i]) %= mod;
}

inline void dfs(int x, int fa = 0) {
    for (int j=0; j<L; ++j) f[x][j] = 0;
    f[x][w[x]] = 1;
    for (int i=head[x]; i; i=nxt[i]) {
        if(to[i] == fa) continue;
        dfs(to[i], x);
        FWT_combine(f[x], f[to[i]]);
    }
    for (int j=0; j<L; ++j) (ans[j] += f[x][j]) %= mod;
}

inline void sol() {
    tot = 0; 
    memset(head, 0, sizeof head);
    memset(ans, 0, sizeof ans);
    cin >> n >> L;
    for (int i=1; i<=n; ++i) scanf("%d", w+i);    
    for (int i=1, u, v; i<n; ++i) {
        scanf("%d%d", &u, &v);
        adde(u, v);
    }
    dfs(1);
    printf("%d", ans[0]);
    for (int i=1; i<L; ++i) printf(" %d", ans[i]);
    puts("");
}

int main() {
    int T; cin >> T;
    inv2 = pwr(2, mod-2);
    while(T--) sol();
    return 0;
}