hdu5909 Tree Cutting（树形dp+fwt_xor优化转移）

Posted 2021-11-24 live4m

题意：

n个点的树,每个点有权值,
对于每个k=[0,m),求有多少个子图满足权值异或和为k.

n<=1e3,m<=1024

解法：

d[i][j]表示以包含点i且以i为根的子树中,异或和为j的方案数.

转移:
d[x][]+=d[x][]卷d[v][]即可.

复杂度O(n^2*log).

ps:
c++编译失败,g++编译成功

code：

#include<bits/stdc++.h>
// #define SYNC_OFF
#define int long long
#define ll long long
#define ull unsigned long long
//fast-coding
#define ST(x) x.begin()
#define ED(x) x.end()
#define RST(x) x.rbegin()
#define RED(x) x.end()
#define CL(x) x.clear();
#define all(a,n) a+1,a+1+n
#define ff(i,n) for(ll i=1;i<=n;i++)
#define rff(i,n) for(ll i=n;i>=1;i--)
#define fff(i,n) for(ll i=0;i<n;i++)
#define rfff(i,n) for(ll i=n-1;i>=0;i--)
#define SC(x) scanf("%s",x)
#define SL(x) strlen(x)
#define pss(a) push_back(a)
#define ps(a) push(a)
#define SZ(x) (int)x.size()
#define pee puts("");
#define eee putchar(' ');
#define re readdd()
#define pr(a) printtt(a)
int readdd(){int x=0,f=1;char c=getchar();//
while(!isdigit(c)&&c!='-')c=getchar();
if(c=='-')f=-1,c=getchar();
while(isdigit(c))x=x*10+c-'0',c=getchar();
return f*x;}
void printtt(int x){if(x<0)putchar('-'),x=-x;//
if(x>=10)printtt(x/10);putchar(x%10+'0');}
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}//
int ppow(int a,int b,int mod){a%=mod;//
int ans=1%mod;while(b){if(b&1)ans=(long long)ans*a%mod;
a=(long long)a*a%mod;b>>=1;}return ans;}
bool addd(int a,int b){return a>b;}
int lowbit(int x){return x&-x;}
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
bool isdigit(char c){return c>='0'&&c<='9';}
bool Isprime(int x){
    for(int i=2;i*i<=x;i++)if(x%i==0)return 0;
    return 1;
}
void ac(int x){if(x)puts("YES");else puts("NO");}
//short_type 
#define VE vector<int>
#define PI pair<int,int>
//
using namespace std;
// const int mod=998244353;
const int mod=1e9+7;
const int maxm=2e6+5;
const int inv2=ppow(2,mod-2,mod);
int d[1111][2222];
int temp[2222];
int ans[2222];
int a[maxm];
VE g[maxm];
int n,m;
/*

*/
/*
*/
int len;
void fwt_xor(int *a, int op){
    for(int i=1;i<len;i<<=1){
        for(int j=0,p=i<<1;j<len;j+=p){
            for(int k=0;k<i;k++){
                int x=a[j+k],y=a[i+j+k];
                a[j+k]=(x+y)%mod;
                a[i+j+k]=(x-y+mod)%mod;
                if(op==-1){
                    a[j+k]=1ll*a[j+k]*inv2%mod;
                    a[i+j+k]=1ll*a[i+j+k]*inv2%mod;
                }
            }
        }
    }
}
void dfs(int x,int fa){
    for(int i=0;i<len;i++)d[x][i]=0;
    d[x][a[x]]=1;
    for(int &v:g[x]){
        if(v==fa)continue;
        dfs(v,x);
        //有卷v和不卷v两种选择
        //将d[x][]复制到temp计算卷的方案数
        for(int i=0;i<len;i++)temp[i]=d[x][i];
        fwt_xor(temp,1);
        fwt_xor(d[v],1);
        for(int i=0;i<len;i++){
            temp[i]=1ll*temp[i]*d[v][i]%mod;
        }
        fwt_xor(temp,-1);
        fwt_xor(d[v],-1);
        //加回d[x][]
        for(int i=0;i<len;i++){
            d[x][i]=(d[x][i]+temp[i])%mod;
        }
    }
}
//c++编译失败,g++编译成功
void solve(){
    n=re,m=re;
    ff(i,n)a[i]=re;
    ff(i,n)CL(g[i]);
    ff(i,n-1){
        int x=re,y=re;
        g[x].pss(y);
        g[y].pss(x);
    }
    //dp
    len=1;
    while(len<m)len<<=1;
    dfs(1,1);
    //
    fff(j,m)ans[j]=0;
    ff(i,n){
        fff(j,m){
            ans[j]=(ans[j]+d[i][j])%mod;
        }
    }
    fff(j,m){
        if(j)eee;
        pr(ans[j]);
    }
    pee;
}
void Main(){
    #define MULTI_CASE
    #ifdef MULTI_CASE
    int T;cin>>T;while(T--)
    #endif
    solve();
}
void Init(){
    #ifdef SYNC_OFF
    ios::sync_with_stdio(0);cin.tie(0);
    #endif
    #ifndef ONLINE_JUDGE
    freopen("../in.txt","r",stdin);
    freopen("../out.txt","w",stdout);
    #endif
}
signed main(){
    Init();
    Main();
    return 0;
}

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