1

ここでは、すべての反復で先行する頂点が表示されます。頂点の最後の先行を表示したい

import java.io.*;
import java.util.*;
public class BellmanFord {
    LinkedList<Edge> edges;
    int d[];
    int n,e,s;
    final int INFINITY=999;
    private static class Edge  {
        int u,v,w;
        public Edge(int a, int b, int c){
            u=a;
            v=b;
            w=c;
       }
       BellmanFord() throws IOException{
            int item;
            edges=new LinkedList<Edge>();
            BufferedReader inp = new BufferedReader (new InputStreamReader(System.in));
            System.out.print("Enter number of vertices ");
            n=Integer.parseInt(inp.readLine());
            System.out.println("Cost Matrix");
            for(int i=0;i<n;i++)
            for(int j=0;j<n;j++){
                item=Integer.parseInt(inp.readLine());
                if(item!=0)
                    edges.add(new Edge(i,j,item));
            }
            e=edges.size();
            d=new int[n];
            System.out.print("Enter the source vertex ");
            s=Integer.parseInt(inp.readLine());
       }
       void relax() {
            int i,j;
            for(i=0;i<n;++i)
                d[i]=INFINITY;;
            d[s] = 0;
            for (i = 0; i < n - 1; ++i)
                 for (j = 0; j < e; ++j)
                     if (d[edges.get(j).u] + edges.get(j).w < d[edges.get(j).v])
                     {                         
                         d[edges.get(j).v] = d[edges.get(j).u] + edges.get(j).w;
                        /*Gives me the predecessor nodes of all iterations How can i get the final predecessornodes*/                                                                                                     System.out.println(edges.get(j).v+" Has predecessor " + edges.get(j).u);
                     }
        }
        boolean cycle() {
            int j;
            for (j = 0; j < e; ++j)
                if (d[edges.get(j).u] + edges.get(j).w < d[edges.get(j).v])
                    return false;
            return true;
        }
        public static void main(String args[]) throws IOException   {
             BellmanFord r=new BellmanFord();
             r.relax();
             if(r.cycle())
             for(int i=0;i<r.n;i++)
             System.out.println(r.s+"to"+i+" ==> "+r.d[i]);
             else
             System.out.println(" There is a nagative edge cycle ");
       }
}

誤った出力は次のとおりです。私はすべての反復について前任者を印刷しようとしています：

**OUTPUT:**

Enter number of vertices
Cost Matrix
0
-1
4
0
0
0
0
3
2
2
0
0
0
0
0
0
1
5
0
0
0
0
0
-3
0
Enter the source vertex
1 Has predecessor 0
2 Has predecessor 0
2 Has predecessor 1
3 Has predecessor 1
4 Has predecessor 1
3 Has predecessor 4
0to0 ==> 0
0to1 ==> -1
0to2 ==> 2
0to3 ==> -2
0to4 ==> 1
4

1 に答える 1

0

あなたの最後の行は非常に理解しにくいようですので、数日前に作成した私のプログラムを紹介します。100行のコードを理解してエラーを見つけるのが難しいため、プログラムが犯す間違いを探すことができます
。時間の最適化にすぐに焦点を合わせるのではなく、きちんとしたコメント付きのコードを書くことにもっと焦点を合わせるようにアドバイスします。ロジックをチェックしてコードに実装してみてください。そのため、すべてを簡単に取得できるようにJavaコードを投稿しませんでした:)

// A C / C++ program for Bellman-Ford's single source shortest path algorithm.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

// a structure to represent a weighted edge in graph
struct Edge
{
    int src, dest, weight;
};

// a structure to represent a connected, directed and weighted graph
struct Graph
{
    // V-> Number of vertices, E-> Number of edges
    int V, E;

    // graph is represented as an array of edges.
    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 utility function used to print the solution
void printArr(int dist[], int n)
{
    printf("Vertex   Distance from Source\n");
    for (int i = 0; i < n; ++i)
        printf("%d \t\t %d\n", i, dist[i]);
}

// The main function that finds shortest distances from src to all other
// vertices using Bellman-Ford algorithm.  The function also detects negative
// weight cycle
void BellmanFord(struct Graph* graph, int src)
{
    int V = graph->V;
    int E = graph->E;
    int dist[V];

    // Step 1: Initialize distances from src to all other vertices as INFINITE
    for (int i = 0; i < V; i++)
        dist[i]   = INT_MAX;
    dist[src] = 0;

    // Step 2: Relax all edges |V| - 1 times. A simple shortest path from src
    // to any other vertex can have at-most |V| - 1 edges
    for (int i = 1; i <= V-1; i++)
    {
        for (int j = 0; j < E; j++)
        {
            int u = graph->edge[j].src;
            int v = graph->edge[j].dest;
            int weight = graph->edge[j].weight;
            if (dist[u] + weight < dist[v])
                dist[v] = dist[u] + weight;
        }
    }

    // Step 3: check for negative-weight cycles.  The above step guarantees
    // shortest distances if graph doesn't contain negative weight cycle.  
    // If we get a shorter path, then there is a cycle.
    for (int i = 0; i < E; i++)
    {
        int u = graph->edge[i].src;
        int v = graph->edge[i].dest;
        int weight = graph->edge[i].weight;
        if (dist[u] + weight < dist[v])
            printf("Graph contains negative weight cycle");
    }

    printArr(dist, V);

    return;
}

// Driver program to test above functions
int main()
{

    int V = 5;  // Number of vertices in graph
    int E = 8;  // 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 = -1;

    // add edge 0-2 
    graph->edge[1].src = 0;
    graph->edge[1].dest = 2;
    graph->edge[1].weight = 4;

    // add edge 1-2 
    graph->edge[2].src = 1;
    graph->edge[2].dest = 2;
    graph->edge[2].weight = 3;

    // add edge 1-3 
    graph->edge[3].src = 1;
    graph->edge[3].dest = 3;
    graph->edge[3].weight = 2;

    // add edge 1-4 
    graph->edge[4].src = 1;
    graph->edge[4].dest = 4;
    graph->edge[4].weight = 2;

    // add edge 3-2 
    graph->edge[5].src = 3;
    graph->edge[5].dest = 2;
    graph->edge[5].weight = 5;

    // add edge 3-1 
    graph->edge[6].src = 3;
    graph->edge[6].dest = 1;
    graph->edge[6].weight = 1;

    // add edge 4-3 
    graph->edge[7].src = 4;
    graph->edge[7].dest = 3;
    graph->edge[7].weight = -3;

    BellmanFord(graph, 0);

    return 0;
}
于 2013-03-27T05:58:36.027 に答える