Warshall‘s algorithm 算法的实现及优化

用 Warshall’s 算法计算传递闭包

(1)时间复杂度为：O(nnn)
(2)用R的无穷闭包时间复杂度为O(nnn*(n - 1))

经典实现：

下面展示一些 经典代码片。

void computeAPSP(const int n) {
    /* calculate shortest paths from every vertex to every vertex */
    for (int k = 0; k < n; k++) 
    {
        for (int i = 0; i < n; i++) 
        {
            for (int j = 0; j < n; j++) 
            {
                a[i][j] = min( a[i][j], a[i][k] + a[k][j] );
            }
        }
    }
}

利用矩阵的对称性优化：

void computeAPSP(const int n) 
{
    for (int k = 0; k < n; k++) 
    {
        for (int i = 0; i < n; i++) 
        {
            if (k != i) 
            {

                const int a_ki = (k < i) ? a[i][k] : a[k][i];

                for (int j = 0; j < min(k, i); j++)
                    a[i][j] = min( a[i][j], a_ki + a[k][j] );

                for (int j = k + 1; j < i; j++)
                    a[i][j] = min( a[i][j], a_ki + a[j][k] );

            }
        }
    }
}

只使用矩阵的下三角部分进行优化：

void computeAPSP(const int n) 
{
    for (int k = 0; k < n; k++) 
    {
        for (int i = 0; i < n; i++) 
        {
            if (k != i) 
            {

                const int a_ki = (k < i) ? a[i][k] : a[k][i];

                for (int j = 0; j < min(k, i); j++)
                    a[i][j] = min( a[i][j], a_ki + a[k][j] );

                for (int j = k + 1; j < i; j++)
                    a[i][j] = min( a[i][j], a_ki + a[j][k] );

            }
        }
    }
}

避免大量调用数学函数进行优化：

void computeAPSP(const int n) 
{
    for (int k = 0; k < n; k++) 
    {
        for (int i = 0; i < n; i++) 
        {
            if (k != i) 
            {

                const int a_ki = (k < i) ? a[i][k] : a[k][i];

                // skip if no path
                if (a_ki == POSITIVE_INFINITY)    continue;

                for (int j = 0; j < min(k, i); j++) 
                {
                    const int s_kj = a_ki + a[k][j];
                    if( s_kj < a[i][j] )    a[i][j] = s_kj;
                }

                for (int j = k + 1; j < i; j++) 
                {
                    const int s_jk = a_ki + a[j][k];
                    if( s_jk < a[i][j] )    a[i][j] = s_jk;
                }

            }
        }
    }
}