[HDU5909]Tree Cutting

Posted 2021-02-11 wolfycz

题目传送门：Vjudge，HDU

题目大意：

给定一棵树，每个节点上有权值，定义一个树上联通块(S)的权值为(V_1oplus V_2oplus...oplus V_k(forall iin S))，问权值为(x(xin[0,m))的联通块个数

---

考虑dp，设(f[i][j])表示以(i)为根，子数中权值为(j)的联通块个数，合并显然就是两个(f)异或卷积一下，FWT优化即可

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
    static char buf[1000000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
    int x=0,f=1;char ch=gc();
    for (;ch<'0'||ch>'9';ch=gc())   if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
    return x*f;
}
inline int read(){
    int x=0,f=1;char ch=getchar();
    for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')    f=-1;
    for (;ch>='0'&&ch<='9';ch=getchar())    x=(x<<1)+(x<<3)+ch-'0';
    return x*f;
}
inline void print(int x){
    if (x<0)    putchar('-'),x=-x;
    if (x>9)    print(x/10);
    putchar(x%10+'0');
}
const int N=1e3,M=1<<10,Mod=1e9+7,inv=5e8+4;
void div(int &x){x=1ll*x*inv%Mod;}
void FWT(int *a,int n,int flag){
    for (int i=2;i<=n;i<<=1){
        for (int j=0;j<n;j+=i){
            for (int k=0;k<i>>1;k++){
                int x=a[j+k],y=a[j+k+(i>>1)];
                a[j+k]=(x+y)%Mod,a[j+k+(i>>1)]=(x-y+Mod)%Mod;
                if (!~flag) div(a[j+k]),div(a[j+k+(i>>1)]);
            }
        }
    }
}
int pre[(N<<1)+10],now[N+10],child[(N<<1)+10];
int f[N+10][M+10],Ans[M+10];
int n,m,tot;
void join(int x,int y){pre[++tot]=now[x],now[x]=tot,child[tot]=y;}
void insert(int x,int y){join(x,y),join(y,x);}
void dfs(int x,int fa){
    FWT(f[x],m,1);
    for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){
        if (son==fa)    continue;
        dfs(son,x);
        for (int i=0;i<m;i++)   f[x][i]=1ll*f[x][i]*f[son][i]%Mod;
    }
    FWT(f[x],m,-1);
    if (++f[x][0]==Mod) f[x][0]=0;
    FWT(f[x],m,1);
}
int main(){
    for (int T=read();T;T--){
        n=read(),m=read(),tot=0;
        memset(f,0,sizeof(f));
        memset(now,0,sizeof(now));
        memset(Ans,0,sizeof(Ans));
        for (int i=1;i<=n;i++)  f[i][read()]=1;
        for (int i=1;i<n;i++){
            int x=read(),y=read();
            insert(x,y);
        }
        dfs(1,0);
        for (int i=1;i<=n;i++)  FWT(f[i],m,-1);
        for (int i=1;i<=n;i++)  if (--f[i][0]<0)    f[i][0]+=Mod;
        for (int i=1;i<=n;i++)
            for (int j=0;j<m;j++)
                Ans[j]=(Ans[j]+f[i][j])%Mod;
        for (int i=0;i<m;i++)   printf("%d",Ans[i]),putchar(i==m-1?'
':' ');
    }
    return 0;
}