HDU - 5909 Tree Cutting (树形dp+FWT优化)

题意：树上每个节点有权值，定义一棵树的权值为所有节点权值异或的值。求一棵树中，连通子树值为[0,m)的个数。

分析：

设\(dp[i][j]\)为根为i,值为j的子树的个数。

则\(dp[i][j\oplus k] = dp[i][j\oplus k] +dp[i][j] * dp[v][k]\) ，但暴力枚举\(dp[i][j] * dp[v][k]\)，每次的复杂度是\(O(M^2)\)的，总的复杂度将是\(O(NM^2)\)，N和M都是1e3,不行。

实际上每次要求的，是个异或的卷积。可以用FWT来将复杂度优化至\(O(NMlogM)\)。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = (1<<10)+5;
typedef long long LL;
const int mod=  1e9+7;
LL dp[1005][MAXN];
LL ans[MAXN];
LL a[1005];
int N,M;
LL qpow(LL a,LL n)
{
    LL res=1;
    while(n){
        if(n &1) res = res* a%mod;
        a = a*a %mod;
        n>>=1;
    }
    return res;
}
LL rev2 = qpow(2,mod-2);

void FWT(LL 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++){
                LL x = a[i+j],y = a[i+j+d];
                a[i+j] = (x+y)%mod,a[i+j+d] = (x-y+mod)%mod;        //取模
            }
        }
    }
}

void UFWT(LL 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++){
                LL x = a[i+j],y = a[i+j+d];
                a[i+j] = (x+y)*rev2%mod,a[i+j+d] = ((x-y)*rev2%mod + mod) % mod;  //取模的情况
            }
        }
    }
}
void solve(LL a[],LL b[],int n){
    FWT(a,n);
    FWT(b,n);
    for (int i = 0 ; i<n ; i++)
        a[i]=a[i]*b[i] %mod;                 //取模
    UFWT(a,n);
}

struct Edge{
    int v,next;
}edges[2005];
int head[1005],tot;
void init(){
    tot = 0;
    memset(head,-1,sizeof(head));
}

void AddEdge(int u,int v)
{
    edges[tot] = (Edge){v,head[u]};
    head[u] = tot++;
}
LL tmp[MAXN];

void dfs(int u,int fa)
{
    dp[u][a[u]] = 1;
    for(int i=head[u];~i;i=edges[i].next){
        int v = edges[i].v;
        if(v== fa) continue;
        dfs(v,u);
        for(int i=0;i<M;++i){
            tmp[i] = dp[u][i];
        }
        solve(dp[u],dp[v],M);
        for(int i=0;i<M;++i){
            dp[u][i] = (dp[u][i] + tmp[i])%mod;     //将之前
        }
    }
    for(int i=0;i<M;++i){
        ans[i] = (ans[i]+ dp[u][i]) %mod;
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int T; scanf("%d",&T);
    while(T--){
        init();
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));
        scanf("%d %d",&N, &M);
        for(int i=1;i<=N;++i){
            scanf("%lld",&a[i]);
        }
        for(int i=1,u,v;i<=N-1;++i){
            scanf("%d %d",&u,&v);
            AddEdge(u,v);
            AddEdge(v,u);
        }
        dfs(1,-1);
        for(int i=0;i<M;++i){
            printf("%lld%c",ans[i],i==M-1?'\n':' ');
        }
    }
    return 0;
}