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

24 42 0 1 31 21 31 44 40 1 3 11 21 31 4

Sample Output

3 3 2 32 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<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<vector>using namespace std;typedef long long LL;const int N=1e3+100; //2^10const 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<int> 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<n;d<<=1)        for(int m=d<<1,i=0;i<n;i+=m)            for(int j=0;j<d;j++){                int x=a[i+j],y=a[i+j+d];                a[i+j]=(x+y)%mod,a[i+j+d]=(x-y+mod)%mod;                //xor:a[i+j]=x+y,a[i+j+d]=(x-y+mod)%mod;                //and:a[i+j]=x+y;                //or:a[i+j+d]=x+y;            }}void UFWT(int a[],int n){    for(int d=1;d<n;d<<=1)        for(int m=d<<1,i=0;i<n;i+=m)            for(int j=0;j<d;j++){                int x=a[i+j],y=a[i+j+d];                a[i+j]=1LL*(x+y)*rev%mod,a[i+j+d]=(1LL*(x-y)*rev%mod+mod)%mod;                //xor:a[i+j]=(x+y)/2,a[i+j+d]=(x-y)/2;                //and:a[i+j]=x-y;                //or:a[i+j+d]=y-x;            }}void solve(int a[],int b[],int n){    FWT(a,n);    FWT(b,n);    for(int i=0;i<n;i++) a[i]=1LL*a[i]*b[i]%mod;    UFWT(a,n);}void dfs(int u,int fa){dp[u][val[u]]=1;//自己与自己异或后为 0 的方案数 for(int i=0;i<V[u].size();i++){int v = V [u][i];if(v==fa) continue;dfs(v,u);for(int i=0;i<m;i++) temp[i]=dp[u][i];solve(dp[u],dp[v],m); //当前dp[u]的所有值与dp[v]的所有值异或的结果for(int i=0;i<m;i++) dp[u][i]=(dp[u][i]+temp[i])%mod;}for(int i=0;i<m;i++) ans[i]=(ans[i]+dp[u][i])%mod;}int main(){int t,u,v; scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(ans,0,sizeof(ans));memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++) V[i].clear(); for(int i=1;i<=n;i++) scanf("%d",&val[i]);for(int i=1;i<n;i++){scanf("%d%d",&u,&v);//addedge(u,v);//addedge(v,u);V[u].push_back(v);V[v].push_back(u);}dfs(1,0);for(int i=0;i<m-1;i++)printf("%d ",ans[i]);printf("%d\n",ans[m-1]);}return 0;}