P5169 xtq的异或和

发表于 | 阅读数

求图上有多少存在多少$(u,v)=x$

考虑树状结构直接$(u,v)$直接走，不会出现多出来的路径造成影响。直接$FWT$所有结点的深度异或。

由于加了个图，只要考虑环的影响，环当作线性基可以取或不取，再做一遍$FWT$即可。

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1 << 18;
const int mod = 1e9 + 7;
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 << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
const int MAXL = 20;
struct LinearBase
{
    ll p[MAXL];

    void clear()
    {
        memset(p, 0, sizeof(p));
    }
    void insert(ll x)
    {
        for (int i = MAXL - 1; i >= 0; i--)
        {
            if (x & (1ll << i))
            {
                if (!p[i])
                {
                    p[i] = x;
                    break;
                }
                x ^= p[i];
            }
        }
    }
    ll query_max(ll x = 0)
    {
        ll res = x;
        for (int i = MAXL - 1; i >= 0; i--)
            res = max(res, res ^ p[i]);
        return res;
    }

    ll query_min()
    {
        for (int i = 0; i < MAXL; i++)
            if (p[i])
                return p[i];
        return 0;
    }
    void rebuild()
    {
        for (int i = MAXL - 1; i >= 0; i--)
            for (int j = i - 1; j >= 0; j--)
                if ((p[i] >> j) & 1)
                    p[i] ^= p[j];
    }
    void mergeFrom(const LinearBase &other)
    {
        for (int i = 0; i <= MAXL; i++)
            insert(other.p[i]);
    }
    ll query_kth(ll k, int n)
    {
        rebuild();
        vector<ll>
            pp;
        for (int i = 0; i < MAXL; ++i)
            if (p[i])
                pp.push_back(p[i]);
        if (pp.size() != n)
            k--;
        if (k > (1LL << pp.size()) - 1)
            return -1;
        ll ans = 0;
        for (int i = 0; i < pp.size(); ++i)
            if (k & (1LL << i))
            {
                ans ^= pp[i];
            }
        return ans;
    }
} lb;
int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}
int inc(int x, int y, int mo)
{
    if (y < 0)
        y += mo;
    if (x + y >= mo)
        x -= mo;
    return x + y;
}
int Inv2;
void FWT(int *A, int n, int op, int t) //t=1 or t=2 and t=3 xor
{
    for (int i = 2; i <= n; i <<= 1)
    {
        for (int j = 0, mid = i >> 1; j < n; j += i)
            for (int k = 0; k < mid; k++)
            {

                if (t == 1)
                    A[j + mid + k] = inc(A[j + mid + k], A[j + k] * op, mod);
                else if (t == 2)
                    A[j + k] = inc(A[j + k], A[j + mid + k] * op, mod);
                else if (t == 3)
                {
                    int x = A[j + k], y = A[j + mid + k];
                    if (op == 1)
                        A[j + k] = (x + y) % mod, A[j + mid + k] = (x - y + mod) % mod;
                    else
                        A[j + k] = 1ll * Inv2 * (x + y) % mod, A[j + mid + k] = 1ll * Inv2 * (x - y + mod) % mod;
                }
            }
    }
}
void FWTX(int *A, int *B, int n, int t)
{
    Inv2 = qpow(2, mod - 2, mod);
    FWT(A, n, 1, t), FWT(B, n, 1, t);

    for (int i = 0; i < n; ++i)
        A[i] = 1ll * A[i] * B[i] % mod;
    FWT(A, n, -1, t);
}
vector<pii> g[N];
int dep[N], vis[N];
int a[N], b[N], c[N];
void work(int x, int d)
{
    if (x == MAXL)
    {
        c[d] = 1;
        return;
    }
    if (lb.p[x])
        work(x + 1, d ^ lb.p[x]);
    work(x + 1, d);
}
void dfs(int x, int d)
{
    dep[x] = d;
    vis[x] = 1;
    for (pii now : g[x])
    {
        int to = now.first;
        if (vis[to])
        {
            lb.insert(dep[to] ^ dep[x] ^ now.second);
        }
        else
        {
            dfs(to, d ^ now.second);
        }
    }
}

int main()
{
    int n = read(), m = read(), q = read();
    for (int i = 1; i <= m; i++)
    {
        int u = read(), v = read(), w = read();
        g[u].push_back(mk(v, w));
        g[v].push_back(mk(u, w));
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++)
    {
        a[dep[i]]++;
        b[dep[i]]++;
    }
    work(0, 0);
    FWTX(a, b, 1 << 18, 3);
    FWTX(a, c, 1 << 18, 3);
    while (q--)
    {
        int x = read();
        printf("%d\n", a[x]);
    }
}