显然的树状DP，dp[u][i]表示以u为根的子树中，价值为i的子树的数量。
转移也很显然。开始时，dp[u][val[u]]为1。每次多一个子树v，就是dp[u]' = dp[u] + ∑dp[u]*dp[v](卷积)。使用FWT处理即可。复杂度O(N*M*logM)
1 #include <cstdio> 2 using namespace std; 3 typedef long long ll; 4 const int MAXN = 1100,MAXM = 2100,mo = 1000000007; 5 int head[MAXN],v[MAXN],to[MAXM],nxt[MAXM],res[1024],dp[MAXN][1024]; 6 int n,m,T,cnt,inv2; 7 int fpow(int x,int k) 8 { 9 if (k == 1) 10 return x; 11 int t = fpow(x,k >> 1); 12 if (k & 1) 13 return (ll)t * t % mo * x % mo; 14 return (ll)t * t % mo; 15 } 16 void add(int x,int y) 17 { 18 nxt[++cnt] = head[x]; 19 to[cnt] = y; 20 head[x] = cnt; 21 } 22 void fwt(int *a,int len,bool inv) 23 { 24 for(int d = 1;d < len;d <<= 1) 25 { 26 for(int h = d << 1,i = 0;i <= len - 1;i += h) 27 for(int j = 0;j < d;j++) 28 { 29 int x = a[i + j],y = a[i + j + d]; 30 a[i + j] = (x + y) % mo; 31 a[i + j + d]=(x - y + mo) % mo; 32 } 33 if (inv == true) 34 for (int i = 0;i <= len - 1;i++) 35 a[i] = (ll)a[i] * inv2 % mo; 36 } 37 } 38 void dfs(int x,int frm) 39 { 40 int tmp1[1024],tmp2[1024]; 41 dp[x][v[x]] = 1; 42 for (int i = head[x];i;i = nxt[i]) 43 { 44 if (to[i] == frm) 45 continue; 46 dfs(to[i],x); 47 for (int o = 0;o <= m - 1;o++) 48 { 49 tmp1[o] = dp[x][o]; 50 tmp2[o] = dp[to[i]][o]; 51 } 52 fwt(tmp1,m,false); 53 fwt(tmp2,m,false); 54 for (int o = 0;o <= m - 1;o++) 55 tmp1[o] = (ll)tmp1[o] * tmp2[o] % mo; 56 fwt(tmp1,m,true); 57 for (int o = 0;o <= m - 1;o++) 58 dp[x][o] = (dp[x][o] + tmp1[o]) % mo; 59 } 60 } 61 int main() 62 { 63 inv2 = fpow(2,mo - 2); 64 for (scanf("%d",&T);T != 0;T--) 65 { 66 cnt = 0; 67 scanf("%d%d",&n,&m); 68 for (int i = 1;i <= n;i++) 69 scanf("%d",&v[i]); 70 int tx,ty; 71 for (int i = 1;i <= n - 1;i++) 72 { 73 scanf("%d%d",&tx,&ty); 74 add(tx,ty); 75 add(ty,tx); 76 } 77 dfs(1,0); 78 for (int i = 1;i <= n;i++) 79 for (int j = 0;j <= m - 1;j++) 80 res[j] = (res[j] + dp[i][j]) % mo; 81 for (int i = 0;i <= m - 2;i++) 82 printf("%d ",res[i]); 83 printf("%d ",res[m - 1]); 84 for (int i = 1;i <= cnt;i++) 85 nxt[i] = 0; 86 for (int i = 1;i <= n;i++) 87 head[i] = 0; 88 cnt = 0; 89 for (int i = 1;i <= n;i++) 90 for (int j = 0;j <= m - 1;j++) 91 dp[i][j] = 0; 92 for (int i = 0;i <= m - 1;i++) 93 res[i] = 0; 94 } 95 return 0; 96 }