HDU 5909 Tree Cutting 动态规划 快速沃尔什变换

Tree Cutting

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5909

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 integer T(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 with m 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

Hint

题意

给你一棵树，然后问你，里面有多少个连通块的异或和为j

题解：

树形DP去做，然后我们考虑转移，用fwt去进行优化就好了

至于为什么这么变换的。。。。

俺也不是很清楚：http://picks.logdown.com/posts/179290-fast-walsh-hadamard-transform

代码

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
const int inv = (mod+1)/2;
const int maxn = 3005;
int n,m,ans[maxn],val[maxn];
vector<int> E[maxn];
long long dp[maxn][maxn],tmp[maxn],a[maxn],b[maxn];
void fwt(long long *aa,int l,int r)
{
    if(r-l==1)return;
    int mid=(l+r)/2;
    fwt(aa,l,mid);
    fwt(aa,mid,r);
    int len=mid-l;
    for(int i=l;i<mid;i++)
    {
        long long x1=aa[i];
        long long x2=aa[i+len];
        aa[i]=(x1+x2)%mod;
        aa[i+len]=(x1-x2+mod)%mod;
    }
}
void ifwt(long long *aa,int l,int r)
{
    if(r-l==1)return;
    int mid=(l+r)/2;
    int len=mid-l;
    for(int i=l;i<mid;i++)
    {
        long long y1=aa[i];
        long long y2=aa[i+len];
        aa[i]=(y1+y2)*inv%mod;
        aa[i+len]=((y1-y2+mod)%mod*inv)%mod;
    }
    ifwt(aa,l,mid);
    ifwt(aa,mid,r);
}
void solve(long long *aa,long long *bb)
{
    memcpy(a,aa,sizeof(long long)*m);
    memcpy(b,bb,sizeof(long long)*m);
    fwt(a,0,m);
    fwt(b,0,m);
    memset(tmp,0,sizeof(tmp));
    for(int i=0;i<m;i++)
        tmp[i]=a[i]*b[i]%mod;
    ifwt(tmp,0,m);
}
void dfs(int x,int f)
{
    memset(dp[x],0,sizeof(dp[x]));
    dp[x][val[x]]=1;
    for(int i=0;i<E[x].size();i++){
        int v=E[x][i];
        if(v==f)continue;
        dfs(v,x);
        solve(dp[x],dp[v]);
        for(int j=0;j<m;j++)(dp[x][j]+=tmp[j])%=mod;
    }
    for(int i=0;i<m;i++)(ans[i]+=dp[x][i])%=mod;
}
void solve()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++)ans[i]=0;
    for(int i=1;i<=n;i++)scanf("%d",&val[i]),E[i].clear();
    for(int i=1;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        E[x].push_back(y);
        E[y].push_back(x);
    }
    dfs(1,0);
    for(int i=0;i<m;i++)
    {
        if(i==0)printf("%d",ans[i]);
        else printf(" %d",ans[i]);
    }
    printf("\n");
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)solve();
    return 0;
}