HDU 5909 Tree Cutting （树形dp+FWT）

Tree Cutting

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 183    Accepted Submission(s): 77

Problem Description

Byteasar has a tree T with n vertices conveniently labeled with 1,2,...,n. Each vertex of the tree has an integer value vi.
The value of a non-empty tree T is equal to v1⊕v2⊕...⊕vn, where ⊕ denotes bitwise-xor.
Now for every integer k from [0,m), please calculate the number of non-empty subtree of T which value are equal to k.
A subtree of T is a subgraph of T that is also a tree.

Input

The first line of the input contains an integerT(1≤T≤10), denoting the number of test cases.
In each test case, the first line of the input contains two integers n(n≤1000) and m(1≤m≤210), denoting the size of the tree T and the upper-bound of v.
The second line of the input contains n integers v1,v2,v3,...,vn(0≤vi<m), denoting the value of each node.

Each of the following n−1 lines contains two integers ai,bi, denoting an edge between vertices ai and bi(1≤ai,bi≤n).

It is guaranteed that m can be represent as 2k, where k is a non-negative integer.

Output

For each test case, print a line withm integers, the i-th number denotes the number of non-empty subtree of T which value are equal to i.
The answer is huge, so please module 109+7.

Sample Input

2 4 4 2 0 1 3 1 2 1 3 1 4 4 4 0 1 3 1 1 2 1 3 1 4

Sample Output

3 3 2 3 2 4 2 3

Source

BestCoder Round #88

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题意：

有一棵n个点的无根树，节点依次编号为1到n，其中节点i的权值为v。
定义一棵树的价值为它所有点的权值的异或和。
现在对于每个[0,m)的整数k，请统计有多少T的非空连通子树的价值等于k。
一棵树T的连通子树就是它的一个连通子图，并且这个图也是一棵树。

题解：设dp[ u ][ i ]表示 u 为根的数，异或后得到 i 的方案数。
转移是个异或卷积的形式，可以用FWT加速计算。
时间复杂度:O(n*m*logm).

那么对于一个状态dp[u][i]我们有加入u的一个子树x之后的新的状态dp[u"][ i ]的值dp[u"][ i ]=dp[u][v]+ (异或卷积)∑（i xor j）dp[u][ i ]*dp[x][ j ]。

AC代码：

#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
const int N=1e3+100; //2^10
const int mod=1e9+7;
const int rev=(mod+1)>>1;
//dp[u][i]表示 u 为根的数，异或后得到 i 的方案数 
int val[N], dp[N][N],ans[N];
int temp[N];
int len;
int n,m;
struct Edge
{
	int from,to;
		
}edge[2*N];

int cnt;
int head[N];
vector V[2*N]; 
//void addedge(int u,int v)
//{
//	edge[cnt].from=v;  edge[cnt].to=head[u];  head[u]=cnt++;
//	edge[cnt].from=u;  edge[cnt].to=head[v];  head[v]=cnt++;
//}

void FWT(int a[],int n)
{
    for(int d=1;d