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    <p>I think your algorithm is mostly correct already, but with slight modifications it becomes trivial to implement.</p> <p>First, <em>every</em> negative edge has to be included in order to minimize the resulting weight. Next, calculate the number of connected components <code>c</code>. If <code>c=1</code>, you're done. Otherwise you need extra <code>c-1</code> positive edges.</p> <p>Now, while you were adding negative edges, consider this already as a Kruskal's algorithm process. Every negative edges may unite a couple of trees in the Kruskal's forest. However, you add the negative edge even if its ends belong to the same tree in the Kruskal's forest — unlike the usual Kruskal's algorithm where you only add those edges that unite two different trees.</p> <p>After this phase, you're left with a graph of <code>c</code> connected components (they may not be trees anymore). Now just continue the Kruskal's algorithm as usual. Process the positive edges in the increasing order, keeping track of the number of unions the you've made with positive edges. Once this number gets to <code>c-1</code>, you're done.</p> <p>By the way, all the process of Kruskal's algorithm can be implemented easily if you represent the forest as <a href="http://en.wikipedia.org/wiki/Disjoint-set_data_structure" rel="noreferrer">disjoint-set data structure</a>. It requires just a couple of lines of code to write, and after that it is trivial to keep track of the number of unions that were made.</p> <hr> <p>Some pseudocode follows:</p> <pre><code>sort(edges); c := n; for edge in edges: if edge.weight &lt; 0: if find(edge.firstEnd) != find(edge.secondEnd): --c; unite(edge.firstEnd, edge.secondEnd); else: if c == 1: break; if find(edge.firstEnd) != find(edge.secondEnd): unite(edge.firstEnd, edge.secondEnd); --c; </code></pre> <p>Here <code>unite</code> and <code>find</code> are the functions of disjoint-set data structure.</p>
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    1. COI think there is a flaw in your algorithm. `every negative edge has to be included in order to minimize the resulting weight`, I don't think it is true because not every negative edges are connected together, and sometimes it needs positive edges to connect them. If those positive edges have large weights, the overall weight maybe even bigger than a single negative edge
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    2. CO@JacksonTale Your primary goal is to make the resulting subgraph connected. If you require positive edges to make your subgraph connected, then you require it anyway, and you have to include them even if they are large. Your secondary goal is minimizing the overall weight. Note that the overall weight need not to be negative, it just has to be minimal. Assume there exists a connected subgraph which does not include all the negative edges from the original graph. Then you can pick any non-included negative edge, add it to the graph, and the overall sum of all the edges included will decrease.
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    3. COSorry, I don't think so. for example, {0, 1} is with weight -20, {2, 3} is with weight -10, if edge {1, 3} has weight of 100, I would rather include {0, 1} only, because the overall weight is -20, right? if I do like you said, then the overall weight is 100-20-10=70.
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