fwt优化+树形DP HDU 5909 

 1 //fwt优化+树形DP HDU 5909
 2 //见官方题解
 3 // BestCoder Round #88 http://bestcoder.hdu.edu.cn/
 4 
 5 #include <bits/stdc++.h>
 6 // #include <iostream>
 7 // #include <cstdio>
 8 // #include <cstdlib>
 9 // #include <algorithm>
10 // #include <vector>
11 // #include <queue>
12 // #include <math.h>
13 using namespace std;
14 #define LL long long
15 typedef pair<int,int> pii;
16 const int inf = 0x3f3f3f3f;
17 const int MOD =1e9+7;
18 const int N =1e3+50;
19 #define clc(a,b) memset(a,b,sizeof(a))
20 const double eps = 1e-8;
21 void fre() {freopen("in.txt","r",stdin);}
22 void freout() {freopen("out.txt","w",stdout);}
23 inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}
24 const int rev = (MOD+1)>>1;
25 int a[N],tem[N];
26 int n,m;
27 vector<int> g[N];
28 int dp[N][N];
29 int ans[N];
30 void FWT(int *a,int n)  {  
31     for(int d=1; d<n; d<<=1)  
32         for(int m=d<<1,i=0; i<n; i+=m)  
33             for(int j=0; j<d; j++)  {  
34                 int x=a[i+j],y=a[i+j+d];  
35                 a[i+j]=(x+y)%MOD,a[i+j+d]=(x-y+MOD)%MOD;   
36             }  
37 }  
38   
39 void UFWT(int *a,int n)  {  
40     for(int d=1; d<n; d<<=1)  
41         for(int m=d<<1,i=0; i<n; i+=m)  
42             for(int j=0; j<d; j++) {  
43                 int x=a[i+j],y=a[i+j+d];  
44                 a[i+j]=1LL*(x+y)*rev%MOD,a[i+j+d]=(1LL*(x-y)*rev%MOD+MOD)%MOD;   
45             }  
46 }  
47   
48 void solve(int *a,int *b,int n)  {  
49     FWT(a,n);  
50     FWT(b,n);  
51     for(int i=0; i<n; i++)   a[i]=1LL*a[i]*b[i]%MOD;  
52     UFWT(a,n);  
53 }  
54 
55 void dfs(int u,int f){
56     dp[u][a[u]]=1;
57     for(int i=0;i<(int)g[u].size();i++){
58         int v=g[u][i];
59         if(v==f) continue;
60         dfs(v,u);
61         for(int j=0;j<m;j++) tem[j]=dp[u][j];
62         solve(dp[u],dp[v],m);
63         for(int j=0;j<m;j++) dp[u][j]=(dp[u][j]+tem[j])%MOD;
64     }
65     for(int i=0;i<m;i++) ans[i]=(ans[i]+dp[u][i])%MOD;
66 }
67 int main(){
68     // fre();
69     int T;
70     scanf("%d",&T);
71     while(T--){
72         clc(dp,0);
73         clc(ans,0);
74         scanf("%d%d",&n,&m);
75         for(int i=1;i<=n;i++) {
76             scanf("%d",&a[i]);
77             g[i].clear();
78         }
79         for(int i=1;i<=n-1;i++){
80             int u,v;
81             scanf("%d%d",&u,&v);
82             g[u].push_back(v);
83             g[v].push_back(u);
84         }
85         dfs(1,0);
86         for(int i=0;i<m;i++){
87             printf("%d%c",ans[i],i==m-1 ? '\n':' ');
88         }
89     }
90     return 0;
91 }

 

posted @ 2016-10-03 16:43  yyblues  阅读(290)  评论(0)    收藏  举报