hdu 5909 Tree Cutting [树形DP fwt]

hdu 5909 Tree Cutting

题意：一颗无根树，每个点有权值，连通子树的权值为异或和，求异或和为[0,m)的方案数

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\(f[i][j]\)表示子树i中经过i的连通子树异或和为j的方案数

转移类似背包，可以用fwt加速

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long ll;

const int N = (1<<10) + 5, P = 1e9+7, inv2 = (P+1)/2;

inline int read() {

    char c=getchar(); int x=0,f=1;

    while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}

    while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}

    return x*f;

}

int n, m, val[N];

struct edge{int v, ne;} e[N<<1];

int cnt, h[N];

inline void ins(int u, int v) {

	e[++cnt] = (edge){v, h[u]}; h[u] = cnt;

	e[++cnt] = (edge){u, h[v]}; h[v] = cnt;

}

int f[N][N], ans[N];

void fwt(int *a, int n, int flag) {

	for(int l=2; l<=n; l<<=1) {

		int m = l>>1;

		for(int *p = a; p != a+n; p += l)

			for(int k=0; k<m; k++) {

				int x = p[k], y = p[k+m];

				if(flag == 1) p[k] = (x + y) %P, p[k+m] = (x - y + P) %P;

				else p[k] = (ll) (x + y) * inv2 %P, p[k+m] = (ll) (x - y + P) * inv2 %P;

			}

	}

}

int t[N];

void dp(int u, int fa) {

	int *a = f[u]; a[val[u]] = 1; //fwt(a, m, 1);

	for(int i=h[u];i;i=e[i].ne) {

		int v = e[i].v;

		if(v == fa) continue;

		dp(v, u); int *b = f[v];

		for(int i=0; i<m; i++) t[i] = a[i];

		fwt(t, m, 1); fwt(b, m, 1);

		for(int i=0; i<m; i++) t[i] = (ll) t[i] * b[i] %P;

		fwt(t, m, -1);

		for(int i=0; i<m; i++) a[i] = (a[i] + t[i]) %P;

	}

	for(int i=0; i<m; i++) ans[i] = (ans[i] + a[i]) %P;

}

int main() {

	freopen("in", "r", stdin);

	int T = read();

	while(T--) {

		n = read(); m = read();

		for(int i=1; i<=n; i++) val[i] = read();

		cnt = 0; memset(h, 0, sizeof(h));

		for(int i=1; i<n; i++) ins(read(), read());

		memset(f, 0, sizeof(f)); memset(ans, 0, sizeof(ans));

		dp(1, 0);

		for(int i=0; i<m; i++) printf("%d%c", ans[i], i==m-1 ? '\n' : ' ');

	}

}