The meaning of problems: the right to give a point value for each tree, the tree is defined as the weight of all the nodes and XOR, interrogates tree have the right subtree of the value of the number k, (0 <= k <m)
First consider the simple algorithm, using dp [i] [j] expressed this point i have is how much XOR sub-tree j, each adding a sub-tree t v tree on the tree right through the weights and v t of enumeration value, updated dp with complexity Om², then the total time complexity is nm²
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 1010; const int maxm = 1200; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; struct Edge{ int to,next; }edge[maxn << 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } int val[maxn]; int dp[maxn][maxm],dp2[maxn][maxm],ans[maxm]; void dfs(int t,int la){ dp[t][val[t]] = 1; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; dfs(v,t); for(int j = 0 ; j < M ; j ++){ for(int k = 0 ; k < M ; k ++){ dp2[t][j ^ k] += dp[t][j] * dp[v][k]; } } for(int j = 0 ; j < M ; j ++){ dp[t][j] += dp2[t][j]; dp2[t][j] = 0; } } for(int i = 0 ; i < M ; i ++) ans[i] += dp[t][i]; } int main(){ int T; Sca(T); while(T--){ Sca2(N,M); init(); for(int i = 0 ; i <= N ; i ++){ for(int j = 0 ; j <= M ; j ++) dp[i][j] = dp2[i][j] = 0; } for(int i = 0 ; i < M; i ++) ans[i] = 0; for(int i = 1; i <= N ; i ++) Sca(val[i]); for(int i = 1; i <= N - 1; i ++){ int u,v; Sca2(u,v); add(u,v); add(v,u); } dfs(1,-1); for(int i = 0 ; i < M; i ++){ printf("%d ",ans[i]); } puts(""); } return 0; }
Then we consider to optimize this layer m², a fact was found that the form 
of the convolution operator logic, can be optimized for the FWT nmlogm, it
Positive Solutions for this problem should be the partition tree, graph theory back up to the tree when the partition code completion, FWT card like a regular card past, cross-GCC C ++ can cross over to TLE
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <bitset> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x) #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x) #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 1010; const int maxm = 1200; const int INF = 0x3f3f3f3f; const LL mod = 1e9 + 7; LL inv2 = mod + 1 >> 1; int N,M,K; struct Edge{ int to,next; }edge[maxn << 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++; } int val[maxn]; LL dp[maxn][maxm],tmp[maxm],ans[maxm]; inline LL add(LL a,LL b){ return ((a + b) % mod + mod) % mod; } inline LL mul(LL a,LL b){ return (a % mod * b % mod + mod) % mod; } void FWT(int limit,LL *a,int op){ for(int i = 1; i < limit; i <<= 1){ for(int p = i << 1,j = 0; j < limit ; j += p){ for(int k = 0 ; k < i; k ++){ LL x = a[j + k],y = a[i + j + k]; a[j + k] = add(x,y); a[i + j + k] = add(x,-y); if(op == -1) a[j + k] = mul(a[j + k],inv2),a[i + j + k] = mul(a[i + j + k],inv2); } } } } void dfs(int t,int la){ dp[t][val[t]] = 1; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; dfs(v,t); for(int j = 0 ; j < M; j ++) tmp[j] = dp[t][j]; FWT(M,tmp,1); FWT(M,dp[v],1); for(int j = 0 ; j < M ; j ++) tmp[j] = mul(tmp[j],dp[v][j]); FWT(M,tmp,-1); for(int j = 0 ; j < M ; j ++) dp[t][j] = add(tmp[j],dp[t][j]); } for(int i = 0 ; i < M ; i ++) ans[i] = add(ans[i],dp[t][i]); } int main(){ int T; Sca(T); while(T--){ Sca2(N,M); init(); for(int i = 0 ; i <= N ; i ++){ for(int j = 0 ; j <= M ; j ++) dp[i][j] = 0; } for(int i = 0; i < M; i ++) ans[i] = tmp[i] = 0; for(int i = 1; i <= N ; i ++) val[i] = read(); for(int i = 1; i <= N - 1; i ++){ int u,v; u = read(); v = read(); add(u,v); add(v,u); } dfs(1,-1); for(int i = 0 ; i < M; i ++) printf("%d%c",ans[i],i == M - 1?'\n':' '); } return 0; }