Question

我之前的问题得到了解答，但我想知道如何改变这一点以显示最低路线经过的交叉路口？感谢

注意：driverMap是一个2D 14x14整数向量，用于保存到达每个顶点所需的距离

void startFirst(vector< vector<int> > driverMap, vector<Car> allCars, int congestFactor)
{
    clock_t start = clock();
    int Intersections[driverMap.size()];
    int Distances[driverMap.size()];

    for(int i = 0; i < driverMap.size(); i++)
    {
        Intersections[i] = i;
    }

    for(int i = 0; i < driverMap.size(); i++)
    {
        cout << "Intersection '" << i << "': "; 
        for(int k = 0; k < driverMap.size(); k++)
        {
            cout << driverMap[i][k] << "|";
        }
        cout << endl;
    }

    for(int i = 0; i < 1; i++)
    {
        int startInt = allCars[i].getStart();
        Intersections[startInt] = -1;
        Distances[startInt] = 0;

        for (int i = 0; i < driverMap.size(); i++) 
        {
            if(i != startInt)
            {
                Distances[i] = driverMap[startInt][i];
            }
        }

        cout << "FOR INTERSECTION: '" << startInt << "'" << endl;
        cout << endl;

        for (int l = 0; l < driverMap.size(); l++) 
        {
            if(l != startInt)
            {
                Dijkstra(driverMap, Intersections, Distances);
            }
        }

        for (int k = 0; k < driverMap.size(); k++) 
        {
            cout << Distances[k] << "|";
        }
    }
    cout << "Total time simulated: " << (clock() - start ) / (double) CLOCKS_PER_SEC << endl;
}

void Dijkstra(vector< vector<int> > driverMap, int Intersections[], int Distances[]) 
{
    int minValue = 9999;
    int minNode = 0;

    for (int i = 0; i < driverMap.size(); i++) 
    {
        if (Intersections[i] == -1) 
        { 
            continue; 
        }

        if (Distances[i] > 0 && Distances[i] < minValue) 
        {
            minValue = Distances[i];
            minNode = i;
        }
    }

    Intersections[minNode] = -1;

    for (int i = 0; i < driverMap.size(); i++) 
    {
        if (driverMap[minNode][i] < 0) 
        { 
            continue; 
        }

        if (Distances[i] < 0) 
        {
            Distances[i] = minValue + driverMap[minNode][i];
            continue;
        }

        if ((Distances[minNode] + driverMap[minNode][i]) < Distances[i]) 
        {
            Distances[i] = minValue + driverMap[minNode][i];
        }
    }
}

Answer 1

下图的手动绘制支持@ hk6279的答案。＆＃34; 0＆＃34;的最佳路径来自＆＃34; 13＆＃34;时无法实现。它应该来自＆＃34; 1＆＃34;或＆＃34; 4＆＃34;到＆＃34; 0＆＃34;并且当@ hk6279注意到并且程序正确计算时，最佳路径是＆＃34; 5＆＃34; - ＆＃34; 8＆＃34; - ＆＃34; 4＆＃34; - ＆＃34; 0＆＃34;距离5200。

enter image description here

Answer 2

您的问题是：假设不正确。 您的假设不正确，因为没有从5到4的直接访问。

您声称可以从5获得0是路径5-4-0，距离4800（2600 + 2200）。但是，事实是您只能通过距离为5-8-4-0（1500 + 1500 + 2200）的代码获取路径5200。

请注意，交叉口5中的距离2600适用于交叉口3，而非交叉口4.您将从索引0开始，而不是1。

Answer 3

请参阅下面的C中Dijkstra算法： http://www.c-program-example.com/2011/10/c-program-to-solve-dijkstras-algorithm.html

Dijkstra算法C ++输出不正确

3 个答案: