digraphs | 易学教程

NetworkX - How to create MultiDiGraph from Shapefile?

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问题 I'm just picking up NetworkX and trying to learn how to use it with Shapefiles. Right now I have a .shp with a road network that I want to represent in a Graph with NetworkX, so that I can find the shortest path between 2 GPS points. I tried using this, but the problem is that when I run the write_shp() function I lose edges because DiGraph does not allow more than one edge between the same two nodes. The arrow in the figure below shows an example of an edge I lose by using a DiGraph. So I

NetworkX - How to create MultiDiGraph from Shapefile?

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Digraph and trigraph can't work together?

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问题 I'm learning digraph and trigraph, and here is the code which I cannot understand. (Yes, I admit that it's extremely ugly.) This code can compile: #define _(s) s%:%:s main(_(_)) <% __; %>t This code can compile, too: #define _(s) s??=??=s main(_(_)) <% __; %> However, neither of the following two pieces of code can compile: #define _(s) s%:??=s main(_(_)) <% __; %> And #define _(s) s??=%:s main(_(_)) <% __; %> This does confuse me: Since the first two pieces of code can compile, I suppose the

What is this smiley-with-beard expression: “<:]{%>”?

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问题 I came across the following program, which compiles without errors or even warnings: int main(){ <:]{%>; // smile! } Live example. What does the program do, and what is that smiley-expression? 回答1: That's an empty lambda using a digraph disguise. Normal lambdas don't have beards. 回答2: The program uses digraphs to represent the following: [] {}; This is a lambda expression that does nothing. The corresponding symbols have these equivalents: <: = [ %> = } Though they are generally unneeded

Colormap not working in NetworkX

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问题 I am trying to make colored graphs. I keep running into odd behavior. I cannot get the colormap to assign colors automatically to the nodes. All nodes I try to do this with wind up the same color! The floats are the colors that are supposed to be assigned to the 6 nodes. Two of the 7 floats are the same because it is a cycle. When I manually specify the color of the nodes ( node_color=['r'] , etc.) it works fine, not only for the root (in red), but for the nodes in the cycle as well. Code: t

Are trigraphs still valid C++?

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问题 We all know about the historical curiosity that is digraphs and trigraphs, but with all the changes made to C++ in recent years I'm curious: are they valid C++14? How about C++17? 回答1: Trigraphs are currently valid, but won't be for long! Trigraphs were proposed for deprecation in C++0x, which was released as C++11. This was opposed by IBM, speaking on behalf of itself and other users of C++, and as a result trigraphs were retained in C++0x. Trigraphs were then proposed again for removal (not

Algorithm for finding shortest cycles in directed graph

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问题 A shortest cycle is one with the minimum number of edges. For example, given a graph: The shortest cycles are: ACDA, DABD If I only needed to find one shortest cycle, I would just run BFS on every vertex and keep track of the smallest cycle. But I don't know how to enumerate all smallest cycles. There is a similar SO question on enumerating minimal cycles in a digraph, but there a minimal cycle is one which is not a union of smaller cycles. Here I am only looking for the cycles with the

What is the >>>= operator in C?

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问题 Given by a colleague as a puzzle, I cannot figure out how this C program actually compiles and runs. What is this >>>= operator and the strange 1P1 literal? I have tested in Clang and GCC. There are no warnings and the output is "???" #include <stdio.h> int main() { int a[2]={ 10, 1 }; while( a[ 0xFULL?'\0':-1:>>>=a<:!!0X.1P1 ] ) printf("?"); return 0; } 回答1: The line: while( a[ 0xFULL?'\0':-1:>>>=a<:!!0X.1P1 ] ) contains the digraphs :> and <: , which translate to ] and [ respectively, so it

What is the >>>= operator in C?

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Given by a colleague as a puzzle, I cannot figure out how this C program actually compiles and runs. What is this >>>= operator and the strange 1P1 literal? I have tested in Clang and GCC. There are no warnings and the output is "???" #include <stdio.h> int main() { int a[2]={ 10, 1 }; while( a[ 0xFULL?'\0':-1:>>>=a<:!!0X.1P1 ] ) printf("?"); return 0; } Ilmari Karonen The line: while( a[ 0xFULL?'\0':-1:>>>=a<:!!0X.1P1 ] ) contains the digraphs :> and <: , which translate to ] and [ respectively, so it's equivalent to: while( a[ 0xFULL?'\0':-1 ] >>= a[ !!0X.1P1 ] ) The literal 0xFULL is the

C++1z why not remove digraphs along with trigraphs?

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问题 C++1z will remove trigraphs. IBM heavily opposed this (here and here) so there seem to be arguments for both sides of removal/non removal. But since the decision was made to remove trigraphs, why leave digraphs? I don't see any reasons for keeping digraphs beyond the reasons to keep trigraphs (which apparently didn't weight enough to keep them). 回答1: Trigraphs are more problematic to the unaware user than digraphs. This is because they are replaced within string literals and comments. Here