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 ? '
' : ' ');
	}
}