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
1 #include <bits/stdc++.h> 2 const int mod = 1e9+7; 3 int rev2 = (mod+1)/2; 4 const double ex = 1e-10; 5 #define inf 0x3f3f3f3f 6 using namespace std; 7 // 下标从0开始 n为2的幂次 8 void FWT(int a[] ,int n){ 9 for (int d = 1 ; d < n ; d <<= 1){ 10 for (int m = d << 1 ,i = 0;i < n ; i+=m){ 11 for (int j = 0 ; j < d ; j++){ 12 int x = a[i+j],y = a[i+j+d]; 13 //xor; 14 a[i+j] = (x+y) % mod,a[i+j+d] = (x-y+mod)%mod; 15 //and 16 //a[i+j]=x+y; 17 //or 18 //a[i+j+d]=x+y; 19 } 20 } 21 } 22 } 23 void UFWT(int a[],int n){ 24 for (int d = 1 ; d < n ; d<<=1){ 25 for (int m = d <<1, i = 0; i < n; i+=m){ 26 for (int j = 0 ; j < d ; j++){ 27 int x = a[i+j],y = a[i+j+d]; 28 //xor 29 a[i+j] = 1LL*(x+y)*rev2%mod,a[i+j+d] = (1LL*(x-y)*rev2%mod + mod) % mod; 30 //and 31 //a[i+j] = x-y; 32 //or 33 //a[i+j+d] = y-x; 34 } 35 } 36 } 37 } 38 void solve(int a[],int b[],int n){ 39 FWT(a,n); 40 FWT(b,n); 41 for (int i = 0 ; i<n ; i++) a[i]=1LL*a[i]*b[i]%mod; 42 UFWT(a,n); 43 } 44 int a[1300]; 45 vector<int> E[1300]; 46 int dp[1300][1300]; 47 int ans[1300]; 48 int tmp[1300]; 49 int m; 50 void dfs(int u,int fa){ 51 dp[u][a[u]] = 1; 52 for (int i = 0 ; i<E[u].size(); i++){ 53 int to = E[u][i]; 54 if (to == fa) continue; 55 dfs(to,u); 56 for (int j = 0;j<m; j++){ 57 tmp[j] = dp[u][j]; 58 } 59 solve(tmp,dp[to],m); 60 for (int j = 0 ; j < m ;j++){ 61 dp[u][j]=(dp[u][j] + tmp[j]) % mod; 62 } 63 } 64 for (int i = 0 ; i < m; i++){ 65 ans[i] = (ans[i] + dp[u][i]) % mod; 66 } 67 } 68 int main() 69 { 70 int t; 71 scanf("%d",&t); 72 while (t--){ 73 int n; 74 scanf("%d%d",&n,&m); 75 for (int i = 1; i<=n;i++) E[i].clear(); 76 for (int i = 1; i<=n ;i++){ 77 scanf("%d",&a[i]); 78 } 79 for (int i=1; i<n ; i++){ 80 int u,v; 81 scanf("%d%d",&u,&v); 82 E[u].push_back(v); 83 E[v].push_back(u); 84 } 85 memset(dp,0,sizeof(dp)); 86 memset(ans,0,sizeof(ans)); 87 dfs(1,0); 88 for (int i = 0; i<m; i++){ 89 printf("%d%c",ans[i],i+1==m?'\n':' '); 90 } 91 } 92 return 0; 93 }
