[HDU5909]Tree Cutting

题目

  点这里看题目。

分析

  不难想到这样的 DP ：
  (f(u,j))：以(u)为根的连通块中异或值(j)的子树的个数。
  转移类似于背包，设已经合并到(u)上的信息为(g)，得到：

[f(u,j)=sum_{(u,v)in E}sum_{ioplus k=j}g(i)f(v,k) ]

  后面这个异或卷积就可以用 FWT 优化了。总时间复杂度(O(Tnmlog_2m))。

代码

#include <cstdio>

const int mod = 1e9 + 7, inv2 = 5e8 + 4;
const int MAXN = 1005, MAXM = ( 1 << 10 ) + 5;

template<typename _T>
void read( _T &x )
{
	x = 0;char s = getchar();int f = 1;
	while( s > '9' || s < '0' ){if( s == '-' ) f = -1; s = getchar();}
	while( s >= '0' && s <= '9' ){x = ( x << 3 ) + ( x << 1 ) + ( s - '0' ), s = getchar();}
	x *= f;
}

template<typename _T>
void write( _T x )
{
	if( x < 0 ){ putchar( '-' ); x = ( ~ x ) + 1; }
	if( 9 < x ){ write( x / 10 ); }
	putchar( x % 10 + '0' );
}

struct edge
{
	int to, nxt;
}Graph[MAXN << 1];

int val[MAXN][MAXM];
int head[MAXN], tot[MAXM];
int N, M, cnt, len;

int fix( const int a ) { return ( a % mod + mod ) % mod; }

void addEdge( const int from, const int to )
{
	Graph[++ cnt].to = to, Graph[cnt].nxt = head[from];
	head[from] = cnt;
}

void FWT( int *f, const int mode )
{
	int t1, t2;
	for( int s = 2 ; s <= len ; s <<= 1 )
		for( int i = 0, t = s >> 1 ; i < len ; i += s )
			for( int j = i ; j < i + t ; j ++ )
			{
				t1 = f[j], t2 = f[j + t];
				if( mode > 0 ) f[j] = ( t1 + t2 ) % mod, f[j + t] = fix( t1 - t2 );
				else f[j] = 1ll * ( t1 + t2 ) * inv2 % mod, f[j + t] = 1ll * fix( t1 - t2 ) * inv2 % mod;
			}
}

void init()
{
	cnt = 0;
	for( int i = 1 ; i <= N ; i ++ ) head[i] = 0;
	for( len = 1 ; len < M ; len <<= 1 ) ;
	for( int i = 1 ; i <= N ; i ++ )
		for( int j = 0 ; j < len ; j ++ )
			val[i][j] = 0;
	for( int i = 0 ; i < M ; i ++ ) tot[i] = 0;
}

void mul( int *ret, int *multi )
{
	for( int i = 0 ; i < len ; i ++ ) 
		ret[i] = 1ll * ret[i] * multi[i] % mod;
}

void DFS( const int u, const int fa )
{
	FWT( val[u], 1 );
	for( int i = head[u], v ; i ; i = Graph[i].nxt )
		if( ( v = Graph[i].to ) ^ fa )
			DFS( v, u ), mul( val[u], val[v] );
	FWT( val[u], -1 );
	( val[u][0] += 1 ) %= mod;
	FWT( val[u], 1 );
}

int main()
{
	int T;
	read( T );
	while( T -- )
	{
		read( N ), read( M ); init();
		for( int i = 1, v ; i <= N ; i ++ ) read( v ), val[i][v] ++;
		for( int i = 1, a, b ; i < N ; i ++ ) read( a ), read( b ), addEdge( a, b ), addEdge( b, a );
		DFS( 1, 0 );
		for( int i = 1 ; i <= N ; i ++ ) FWT( val[i], -1 ), val[i][0] = fix( val[i][0] - 1 );
		for( int i = 1 ; i <= N ; i ++ )
			for( int j = 0 ; j < M ; j ++ ) 
				( tot[j] += val[i][j] ) %= mod;
		for( int i = 0 ; i < M ; i ++ ) write( tot[i] ), putchar( i == M - 1 ? '
' : ' ' );
	}
	return 0;
}