代码搜索:minimum

找到约 5,594 项符合「minimum」的源代码

代码结果 5,594
www.eeworm.com/read/168845/5432676

cpp minimum_degree_ordering.cpp

//-*-c++-*- //======================================================================= // Copyright 1997-2001 University of Notre Dame. // Authors: Lie-Quan Lee // // Distributed under the Boost S
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m minimum_spanning_tree.m

function A = minimum_spanning_tree(C1, C2) % % Find the minimum spanning tree using Prim's algorithm. % C1(i,j) is the primary cost of connecting i to j. % C2(i,j) is the (optional) secondary cost of
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w minimum_degree_ordering.w

\documentclass[11pt]{report} \input{defs} \setlength\overfullrule{5pt} \tolerance=10000 \sloppy \hfuzz=10pt \makeindex \begin{document} \title{An Implementation of the Multiple Minimum Degree Al
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cpp minimum_degree_ordering.cpp

//-*-c++-*- //======================================================================= // Copyright 1997-2001 University of Notre Dame. // Authors: Lie-Quan Lee // // Distributed under the Boost Softwa
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cxx test_bracket_minimum.cxx

#include #include #include #include struct bm_square1 : public vnl_cost_function { bm_square1()
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h vnl_bracket_minimum.h

// This is core/vnl/algo/vnl_bracket_minimum.h #ifndef vnl_bracket_minimum_h_ #define vnl_bracket_minimum_h_ #ifdef VCL_NEEDS_PRAGMA_INTERFACE #pragma interface #endif //: // \file // \brief F
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cxx vnl_bracket_minimum.cxx

// This is core/vnl/algo/vnl_bracket_minimum.cxx #ifdef VCL_NEEDS_PRAGMA_INTERFACE #pragma implementation #endif //: // \file // \brief Function to bracket a minimum // \author Tim Cootes // \
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cpp 1394 minimum inversion number.cpp

#include using namespace std; int num[5001]; int imin,n; int main() { int i,j,t; while( scanf("%d",&n)!=EOF )//输入的数字范围 0 ~ n-1 { for(i=0;i
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hpp prim_minimum_spanning_tree.hpp

//======================================================================= // Copyright 1997, 1998, 1999, 2000 University of Notre Dame. // Authors: Andrew Lumsdaine, Lie-Quan Lee, Jeremy G. Siek // //
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cpp minimum inversion number(逆序数).cpp

#include #include #include using namespace std; const int MAX = 5100; //逆序数值存放在anti中 int num[MAX]; int p[MAX], t[MAX], anti = 0; void merge(int first, int last)