C++ 如何实现双倍权重(而不是整数)的克鲁斯卡尔定理

Question

我在这里找到了一个使用Kruskal算法的最小生成树模板

他们使用整数权重，如果我想使用双倍权重来实现代码，这可能吗？

我在这里和那里做出了改变，并不断给我错误。

以下是我更改的内容：

struct Edge
{
    int src, dest;
    double weight;
};

和

   double myComp(const void* a, const void* b)
    {
        struct Edge* a1 = (struct Edge*)a;
        struct Edge* b1 = (struct Edge*)b;
        return a1->weight > b1->weight;
    }

我不知道为什么，但这些更改使void KruskalMST(struct Graph* graph)中的快速排序在接下来的几行中无法工作

这是原始代码：

// Kruskal's algortihm to find Minimum Spanning Tree of a given connected,
// undirected and weighted graph
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// a structure to represent a weighted edge in graph
struct Edge
{
    int src, dest, weight;
};
// a structure to represent a connected, undirected and weighted graph
struct Graph
{
    // V-> Number of vertices, E-> Number of edges
    int V, E;
    // graph is represented as an array of edges. Since the graph is
    // undirected, the edge from src to dest is also edge from dest
    // to src. Both are counted as 1 edge here.
    struct Edge* edge;
};
// Creates a graph with V vertices and E edges
struct Graph* createGraph(int V, int E)
{
    struct Graph* graph = (struct Graph*) malloc( sizeof(struct Graph) );
    graph->V = V;
    graph->E = E;
    graph->edge = (struct Edge*) malloc( graph->E * sizeof( struct Edge ) );
    return graph;
}
// A structure to represent a subset for union-find
struct subset
{
    int parent;
    int rank;
};
// A utility function to find set of an element i
// (uses path compression technique)
int find(struct subset subsets[], int i)
{
    // find root and make root as parent of i (path compression)
    if (subsets[i].parent != i)
        subsets[i].parent = find(subsets, subsets[i].parent);
    return subsets[i].parent;
}
// A function that does union of two sets of x and y
// (uses union by rank)
void Union(struct subset subsets[], int x, int y)
{
    int xroot = find(subsets, x);
    int yroot = find(subsets, y);
    // Attach smaller rank tree under root of high rank tree
    // (Union by Rank)
    if (subsets[xroot].rank < subsets[yroot].rank)
        subsets[xroot].parent = yroot;
    else if (subsets[xroot].rank > subsets[yroot].rank)
        subsets[yroot].parent = xroot;
    // If ranks are same, then make one as root and increment
    // its rank by one
    else
    {
        subsets[yroot].parent = xroot;
        subsets[xroot].rank++;
    }
}
// Compare two edges according to their weights.
// Used in qsort() for sorting an array of edges
int myComp(const void* a, const void* b)
{
    struct Edge* a1 = (struct Edge*)a;
    struct Edge* b1 = (struct Edge*)b;
    return a1->weight > b1->weight;
}
// The main function to construct MST using Kruskal's algorithm
void KruskalMST(struct Graph* graph)
{
    int V = graph->V;
    struct Edge result[V];  // Tnis will store the resultant MST
    int e = 0;  // An index variable, used for result[]
    int i = 0;  // An index variable, used for sorted edges
    // Step 1:  Sort all the edges in non-decreasing order of their weight
    // If we are not allowed to change the given graph, we can create a copy of
    // array of edges
    qsort(graph->edge, graph->E, sizeof(graph->edge[0]), myComp);
    // Allocate memory for creating V ssubsets
    struct subset *subsets =
        (struct subset*) malloc( V * sizeof(struct subset) );
    // Create V subsets with single elements
    for (int v = 0; v < V; ++v)
    {
        subsets[v].parent = v;
        subsets[v].rank = 0;
    }
    // Number of edges to be taken is equal to V-1
    while (e < V - 1)
    {
        // Step 2: Pick the smallest edge. And increment the index
        // for next iteration
        struct Edge next_edge = graph->edge[i++];
        int x = find(subsets, next_edge.src);
        int y = find(subsets, next_edge.dest);
        // If including this edge does't cause cycle, include it
        // in result and increment the index of result for next edge
        if (x != y)
        {
            result[e++] = next_edge;
            Union(subsets, x, y);
        }
        // Else discard the next_edge
    }
    // print the contents of result[] to display the built MST
    printf("Following are the edges in the constructed MSTn");
    for (i = 0; i < e; ++i)
        printf("%d -- %d == %dn", result[i].src, result[i].dest,
                                                   result[i].weight);
    return;
}
// Driver program to test above functions
int main()
{
    /* Let us create following weighted graph
             10
        0--------1
        |       |
       6|   5   |15
        |       |
        2--------3
            4       */
    int V = 4;  // Number of vertices in graph
    int E = 5;  // Number of edges in graph
    struct Graph* graph = createGraph(V, E);

    // add edge 0-1
    graph->edge[0].src = 0;
    graph->edge[0].dest = 1;
    graph->edge[0].weight = 10;
    // add edge 0-2
    graph->edge[1].src = 0;
    graph->edge[1].dest = 2;
    graph->edge[1].weight = 6;
    // add edge 0-3
    graph->edge[2].src = 0;
    graph->edge[2].dest = 3;
    graph->edge[2].weight = 5;
    // add edge 1-3
    graph->edge[3].src = 1;
    graph->edge[3].dest = 3;
    graph->edge[3].weight = 15;
    // add edge 2-3
    graph->edge[4].src = 2;
    graph->edge[4].dest = 3;
    graph->edge[4].weight = 4;
    KruskalMST(graph);
    return 0;
}

Answer 1

该问题与double与int完全无关，它可能是由于运气好而与int一起使用的，但同样不正确。

int myComp(const void* a, const void* b)
{
    struct Edge* a1 = (struct Edge*)a;
    struct Edge* b1 = (struct Edge*)b;
    return a1->weight > b1->weight;  <----THIS
}

如果a1.weight大于b1.weight，则返回1，否则返回0。

您必须返回的是：

• 0：如果两个值都等于

• <0：如果a1.weight小于b1.weight

• >0：如果a1.weight大于b1.weight

显然，您的函数并没有像qsort所期望的那样运行。更改代码以匹配这些值。你可以查看你提供给我的链接，看看这个例子。