HDU-5909 Tree Cutting
Description
Byteasar has a tree \(T\) with \(n\) vertices conveniently labeled with \(1,2,...,n\). Each vertex of the tree has an integer value \(v_i\).
The value of a non-empty tree \(T\) is equal to \(v_1\oplus v_2\oplus ...\oplus v_n\), where \(\oplus\) denotes bitwise-xor.
Now for every integer \(k\) from \([0,m)\), please calculate the number of non-empty subtree of \(T\) which value are equal to \(k\).
A subtree of \(T\) is a subgraph of \(T\) that is also a tree.
Input
The first line of the input contains an integer \(T(1\leq T\leq10)\), denoting the number of test cases.
In each test case, the first line of the input contains two integers \(n(n\leq 1000)\) and \(m(1\leq m\leq 2^{10})\), denoting the size of the tree \(T\) and the upper-bound of \(v\).
The second line of the input contains \(n\) integers \(v_1,v_2,v_3,...,v_n(0\leq v_i < m)\), denoting the value of each node.
Each of the following \(n-1\) lines contains two integers \(a_i,b_i\), denoting an edge between vertices \(a_i\) and \(b_i(1\leq a_i,b_i\leq n)\).
It is guaranteed that \(m\) can be represent as \(2^k\), where \(k\) is a non-negative integer.
Output
For each test case, print a line with \(m\) integers, the \(i\)-th number denotes the number of non-empty subtree of \(T\) which value are equal to \(i\).
The answer is huge, so please module \(10^9+7\).
Sample Input
2
4 4
2 0 1 3
1 2
1 3
1 4
4 4
0 1 3 1
1 2
1 3
1 4
Sample Output
3 3 2 3
2 4 2 3
题意
求一颗树中异或值为\([0,m)\)的联通块数量
题解
我们先考虑暴力.
设\(f[u][j]\)为必须选择u结点,异或和为j的方案数,那么有\(f[u][i\oplus j]=f[u][i]*f[v][j]\),由于是求所有联通块,而这个答案其实是选择v的联通块的方案数,所以我们还需要把之前的方案数再加回来,也就是
而中间的\(f[u][i\oplus j]=f[u][i]*f[v][j]\)可以使用fwt优化,从而使\(n^2\)的转移降到\(nlogn\),这样总复杂度\(n^2logn\),可以通过
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
ll qpow(ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
const ll inv2 = qpow(2, mod - 2);
void fwt_xor(ll a[], int len, int op) {
for (int h = 2; h <= len; h <<= 1) {
for (int j = 0; j < len; j += h) {
for (int k = j; k < j + h / 2; k++) {
ll u = a[k], t = a[k + h / 2];
a[k] = (u + t) % mod;
a[k + h / 2] = (u - t + mod) % mod;
if (op == -1) {
a[k] = a[k] * inv2 % mod;
a[k + h / 2] = a[k + h / 2] * inv2 % mod;
}
}
}
}
}
const int N = 2050;
vector<int> G[N];
ll f[N][N * 2];
ll tmp[N * 2];
int n, m;
int c[N];
ll ans[N * 2];
void dfs(int u, int fa) {
f[u][c[u]] = 1;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v == fa) continue;
dfs(v, u);
for (int j = 0; j < m; j++) {
tmp[j] = f[u][j];
}
fwt_xor(f[u], m, 1);
fwt_xor(f[v], m, 1);
for (int j = 0; j < m; j++) {
f[u][j] = f[u][j] * f[v][j] % mod;
}
fwt_xor(f[u], m, -1);
for (int j = 0; j < m; j++) {
f[u][j] = (f[u][j] + tmp[j]) % mod;
}
}
for (int j = 0; j < m; j++) {
ans[j] = (ans[j] + f[u][j]) % mod;
}
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
for (int j = 0; j < m; j++) {
f[i][j] = 0;
ans[j] = 0;
}
}
for (int i = 1; i <= n; i++) {
scanf("%d", &c[i]);
G[i].clear();
}
for (int i = 1; i < n; i++) {
int u, v;
scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1, 0);
for (int i = 0; i < m - 1; i++) {
printf("%lld ", ans[i]);
}
printf("%lld\n", ans[m - 1]);
}
return 0;
}
