FWT（Fast-Walsh-hadamard-Transform）模板+例题

看了大神的博客

http://blog.csdn.net/xuanandting/article/details/70991372 除了排版累死人，别的讲的不错

抄了大神的模板，其实就是递归，只不过用for写的，类似于区间DP。复杂度nlogn

// 下标从0开始 n为2的幂次
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];
                //xor;
                a[i+j] = (x+y) % mod,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];
                //xor
                a[i+j] = 1LL*(x+y)*rev2%mod,a[i+j+d] = (1LL*(x-y)*rev2%mod + mod) % mod;
                //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);
}

又看了一个例题

HDU 5909 这种树DP很常见，如果都是异或值可以这么优化，还有的需要类似FFT的优化。

很裸的DP

#include <bits/stdc++.h>
const int mod = 1e9+7;
int rev2 = (mod+1)/2;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
// 下标从0开始 n为2的幂次
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];
                //xor;
                a[i+j] = (x+y) % mod,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];
                //xor
                a[i+j] = 1LL*(x+y)*rev2%mod,a[i+j+d] = (1LL*(x-y)*rev2%mod + mod) % mod;
                //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);
}
int a[1300];
vector<int> E[1300];
int dp[1300][1300];
int ans[1300];
int tmp[1300];
int m;
void dfs(int u,int fa){
    dp[u][a[u]] = 1;
    for (int i = 0 ; i<E[u].size(); i++){
        int to = E[u][i];
        if (to == fa) continue;
        dfs(to,u);
        for (int j = 0;j<m; j++){
            tmp[j] = dp[u][j];
        }
        solve(tmp,dp[to],m);
        for (int j = 0 ; j  < m ;j++){
            dp[u][j]=(dp[u][j] + tmp[j]) % mod;
        }
    }
    for (int  i = 0 ; i < m; i++){
        ans[i] = (ans[i] + dp[u][i]) % mod;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while (t--){
        int n;
        scanf("%d%d",&n,&m);
        for (int i = 1; i<=n;i++) E[i].clear();
        for (int i = 1; i<=n ;i++){
            scanf("%d",&a[i]);
        }
        for (int i=1; i<n ; i++){
            int u,v;
            scanf("%d%d",&u,&v);
            E[u].push_back(v);
            E[v].push_back(u);
        }
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));
        dfs(1,0);
        for (int i = 0; i<m; i++){
            printf("%d%c",ans[i],i+1==m?'
':' ');
        }
    }
    return 0;
}