abc258 G - Triangle

题意：

给定图的邻接矩阵，问三元环的数量

\(n\le 3000\)

思路：

\(O(n^3)\) 的最naive的暴力，用 bitset 优化到 \(O(n^3/64)\)

暴力搜 \((i,j)\)，则另外两个点 \((i,k),(j,k)\) 在同一列

Since \(\frac{3000^3}{64}=421875000\) \(\simeq\) \(4\) \(\times\) \(10^8\), this solution is fast enough. We may further halve the execution time by only searching \((i,j)\) such that \(i< j\).

// 850ms
void sol() {
    int n; cin >> n;
    vector<bitset<3000> > a(n);
    for(int i = 0; i < n; i++) {
        string s; cin >> s;
        for(int j = 0; j < n; j++)
            a[i][j] = s[j] == '1';
    }
    long long ans = 0;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < i; j++)
            if(a[i][j]) ans += (a[i] & a[j]).count();
    cout << ans / 3;
}