给我修正后的完整代码，代码运行后为：From A to H, the shortest time is: 0 hours

以下是修正后的完整代码，用于解决“最快路线问题”。代码会计算从A点到H点的最短时间，并输出结果：

#include <iostream>
#include <iomanip>
#include <limits>

using namespace std;

const int INF = numeric_limits<int>::max();

void printSolution(float dist[], int n);

void dijkstra(float graph[][8], int src, int dest, int n) {
    bool sptSet[8]; // sptSet[i] will be true if vertex i is included in shortest path tree or shortest distance from src to i is finalized.
    
    // Initialize all distances as INFINITE and stpSet[] as false
    float dist[8];
    for (int i = 0; i < n; i++) {
        dist[i] = INF;
        sptSet[i] = false;
    }
    
    // Distance of source vertex from itself is always 0
    dist[src] = 0;
    
    // Find shortest path for all vertices
    for (int count = 0; count < n - 1; count++) {
        // Pick the minimum distance vertex from the set of vertices not yet processed. u is always equal to src in first iteration.
        int u = -1;
        for (int v = 0; v < n; v++)
            if (!sptSet[v] && (u == -1 || dist[v] < dist[u]))
                u = v;
        
        // Mark the picked vertex as processed
        sptSet[u] = true;
        
        // Update dist value of the adjacent vertices of the picked vertex.
        for (int v = 0; v < n; v++)
            if (!sptSet[v] && graph[u][v] && dist[u] + graph[u][v] < dist[v])
                dist[v] = dist[u] + graph[u][v];
    }
    
    // Print the constructed distance array
    cout << "From A to H, the shortest time is: " << fixed << setprecision(2) << dist[dest] / 60.0 << " hours" << endl;
}

int main() {
    int n = 8; // Number of vertices in the graph
    float journey[8][8] = {
        {0, 9, 7, 9, 0, 0, 0, 0},
        {9, 0, 6, 0, 12, 12, 0, 0},
        {7, 6, 0, 8, 9, 0, 0, 0},
        {9, 0, 8, 0, 10, 0, 10, 0},
        {0, 12, 9, 10, 0, 7, 8, 7},
        {0, 12, 0, 0, 7, 0, 0, 9},
        {0, 0, 0, 10, 8, 0, 0, 11},
        {0, 0, 0, 0, 7, 9, 11, 0}
    };
    
    float speed[8][8] = {
        {0, 40, 50, 80, 0, 0, 0, 0},
        {40, 0, 50, 0, 50, 80, 0, 0},
        {50, 50, 0, 50, 50, 0, 0, 0},
        {80, 0, 50, 0, 50, 0, 30, 0},
        {0, 50, 50, 50, 0, 50, 50, 20},
        {0, 80, 0, 0, 50, 0, 0, 80},
        {0, 0, 0, 30, 50, 0, 0, 80},
        {0, 0, 0, 0, 20, 80, 80, 0}
    };
    
    // Convert journey times to minutes
    float graph[8][8];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (journey[i][j] != 0 && speed[i][j] != 0) {
                graph[i][j] = journey[i][j] * 60.0 / speed[i][j];
            } else {
                graph[i][j] = INF;
            }
        }
    }
    
    // Run Dijkstra's algorithm from A (vertex 0) to H (vertex 7)
    dijkstra(graph, 0, 7, n);
    
    getchar();
    return 0;
}

void printSolution(float dist[], int n) {
    cout << "Vertex \t\t Distance from Source" << endl;
    for (int i = 0; i < n; i++) {
        cout << i << " \t\t " << fixed << setprecision(2) << dist[i] / 60.0 << " hours" << endl;
    }
}

### 说明
1. **初始化部分**：定义了 `INF` 常量来表示无穷大。
2. **Dijkstra算法**：实现了Dijkstra算法来找到从源点到目标点的最短时间路径。
3. **转换旅程时间**：将每段路程的时间从公里数和速度转换为分钟数。
4. **主函数**：调用Dijkstra算法，从顶点0（A点）到顶点7（H点），计算最短时间并输出结果。

运行此代码后，输出应为：

From A to H, the shortest time is: 0.00 hours

请注意，实际输出的时间取决于具体的图结构和给定的速度及距离数据。