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Luogu P3298 [SDOI2013]泉

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题目描述

https://www.luogu.com.cn/problem/P3298

Solution

我们考虑组合容斥

令 $f[k]=\sum_{i=k}^6\binom{i}{k}g_k$，我们发现这个东西比较容易计算

我们考虑枚举位置，然后求出这些位置相同的对数，这个东西就是 $f$，然后二项式反演求得 $g$ 即可

时间复杂度 $O(6\cdot 2^6 n+2^6n\log n)$，不用 $hash$ 好像怎么都过不了 = =

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#include <iostream>
#include <cstdio>
#include <vector>
#include <map>
#include <algorithm>
#define maxn 1000010
#define ll long long
using namespace std;

const int N = 6; 

int n, k, a[maxn][N];

int C[10][10]; 
void init_C(int n) {
    for (int i = 0; i <= n; ++i) C[i][0] = 1;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= i; ++j) C[i][j] = C[i - 1][j] + C[i - 1][j - 1]; 
}

const int base = 233333;
const ll p = 212370440130137957;
inline ll mul(ll x, ll y) {
    return (x * y - (ll) ((long double) x / p * y) * p + p) % p;
}

ll Hash(vector<int> &a) {
    ll s = 0;
    for (auto u : a)
        s = (mul(s, base) + u) % p; 
    return s; 
} 

ll f[maxn], g[maxn]; 
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> k;
    for (int i = 1; i <= n; ++i)
        for (int j = 0; j < N; ++j) cin >> a[i][j];
    int M = (1 << N) - 1; 
    for (int S = 0; S <= M; ++S) {
        vector<ll> V;
        int v = __builtin_popcount(S);
        if (v < k) continue;
        for (int i = 1; i <= n; ++i) {
            vector<int> A;
            for (int o = 0; o < N; ++o) 
                if (S >> o & 1) A.push_back(a[i][o]);
            V.push_back(Hash(A)); 
        } sort(V.begin(), V.end()); 
        for (int i = 0, sz = V.size(), p = i; i < sz; i = p) {
            while (p < sz && V[p] == V[i]) ++p;
            f[v] += (ll) (p - i) * (p - i - 1) / 2; 
        } 
    } init_C(6);
    for (int i = k, t = 1; i <= N; ++i, t = -t) g[k] += t * C[i][k] * f[i];
    cout << g[k] << "\n"; 
    return 0;
}