#include #include #include "matrix.h" using namespace std; const int INFINITY = INT_MAX; /** * Compute optimal ordering of matrix multiplication. * c contains the number of columns for each of the n matrices. * c[ 0 ] is the number of rows in matrix 1. * The minimum number of multiplications is left in m[ 1 ][ n ]. * Actual ordering is computed via another procedure using lastChange. * m and lastChange are indexed starting at 1, instead of 0. * Note: Entries below main diagonals of m and lastChange * are meaningless and uninitialized. */ void optMatrix( const vector & c, matrix & m, matrix & lastChange ) { int n = c.size( ) - 1; for( int left = 1; left <= n; ++left ) m[ left ][ left ] = 0; for( int k = 1; k < n; ++k ) // k is right - left for( int left = 1; left <= n - k; ++left ) { // For each position int right = left + k; m[ left ][ right ] = INFINITY; for( int i = left; i < right; ++i ) { int thisCost = m[ left ][ i ] + m[ i + 1 ][ right ] + c[ left - 1 ] * c[ i ] * c[ right ]; if( thisCost < m[ left ][ right ] ) // Update min { m[ left ][ right ] = thisCost; lastChange[ left ][ right ] = i; } } } } int main( ) { vector c = { 50, 10, 40, 30, 5 }; matrix m( 5, 5 ); matrixlastChange( 5, 5 ); optMatrix( c, m, lastChange ); int i; for( i = 1; i < m.numrows( ); ++i ) { for( int j = 1; j < m.numcols( ); ++j ) cout << m[ i ][ j ] << " "; cout << endl; } for( i = 1; i < lastChange.numrows( ); ++i ) { for( int j = 1; j < lastChange.numcols( ); ++j ) cout << lastChange[ i ][ j ] << " "; cout << endl; } return 0; }