题目链接
题解
设(f[i][j])表示以(i)为根的子树，(i)一定取，剩余节点必须联通，异或和为(j)的方案数
初始化(f[i][val[i]] = 1)
枚举儿子(v)转移
[f[i][j] = f[i][j] + sumlimits_{x ; xor ; y = j} f[i][x] centerdot f[v][y]
]
发现是一个异或卷积的形式
前面相当于和(1)的卷积，后面是和(f[v])的卷积
(FWT)即可优化到(O(nlogn + nm))
我被卡常了QAQ，需要注意一些细节
单独一个(1)做(FWT)是全(1)，转移乘(f[v][j])时(+1)即可
每个(f[i])做完(FWT)后不立即(IFWT)，可以直接参与转移，全部做完后再(IFWT)
循环展开大法好
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define Redge(u) for (register int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 2050,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
const int P = 1000000007;
int n,m,deg,val[maxn],A[maxn][maxn],B[maxn],inv2,fa[maxn];
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
inline int qpow(int a,int b){
int re = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) re = 1ll * re * a % P;
return re;
}
inline void fwt(int* a,int n,int f){
for (int i = 1; i < n; i <<= 1)
for (int j = 0; j < n; j += (i << 1))
for (int k = 0; k < i; k++){
int x = a[j + k],y = a[j + k + i];
a[j + k] = (x + y) % P,a[j + k + i] = (x - y + P) % P;
if (f == -1) a[j + k] = 1ll * a[j + k] * inv2 % P,a[j + k + i] = 1ll * a[j + k + i] * inv2 % P;
}
}
void dfs(int u){
for (register int i = 0; i < deg; i += 4){
A[u][i] = 0; A[u][i + 1] = 0;
A[u][i + 2] = 0; A[u][i + 3] = 0;
}
A[u][val[u]] = 1; fwt(A[u],deg,1);
Redge(u) if ((to = ed[k].to) != fa[u]){
fa[to] = u; dfs(to);
for (register int i = 0; i < deg; i++)
A[u][i] = 1ll * A[u][i] * (A[to][i] + 1) % P;
}
}
int main(){
int T = read(); inv2 = qpow(2,P - 2);
while (T--){
n = read(); m = read(); deg = 1; ne = 1;
REP(i,n) h[i] = 0;
while (deg < m) deg <<= 1;
REP(i,n) val[i] = read();
for (int i = 1; i < n; i++) build(read(),read());
dfs(1);
REP(i,n) fwt(A[i],deg,-1);
for (register int i = 0; i < m; i++){
int ans = 0;
for (register int j = 1; j <= n; j++) ans = (ans + A[j][i]) % P;
printf("%d",ans);
if (i != m - 1) putchar(' ');
}
puts("");
}
return 0;
}