hdu5909 Tree Cutting 【树形dp + FWT】

题目链接

hdu5909

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

\[f[i][j] = f[i][j] + \sum\limits_{x \; xor \; y = j} f[i][x] \centerdot f[v][y] \]

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;
}