用c++实现Dijkstra算法求源点V0到其它顶点的最短路径长度

以下是一个使用C++实现Dijkstra算法的示例代码：

#include <iostream>
#include <climits>
using namespace std;

#define V 9

int minDistance(int dist[], bool visited[]) {
    int min = INT_MAX, min_index;
    for (int v = 0; v < V; v++) {
        if (!visited[v] && dist[v] <= min) {
            min = dist[v];
            min_index = v;
        }
    }
    return min_index;
}

void printSolution(int dist[]) {
    cout << "顶点\t最短路径长度" << endl;
    for (int i = 0; i < V; i++) {
        cout << i << "\t" << dist[i] << endl;
    }
}

void dijkstra(int graph[V][V], int src) {
    int dist[V];
    bool visited[V];

    for (int i = 0; i < V; i++) {
        dist[i] = INT_MAX;
        visited[i] = false;
    }

    dist[src] = 0;

    for (int count = 0; count < V - 1; count++) {
        int u = minDistance(dist, visited);
        visited[u] = true;

        for (int v = 0; v < V; v++) {
            if (!visited[v] && graph[u][v] && dist[u] != INT_MAX && dist[u] + graph[u][v] < dist[v]) {
                dist[v] = dist[u] + graph[u][v];
            }
        }
    }

    printSolution(dist);
}

int main() {
    int graph[V][V] = {
        {0, 4, 0, 0, 0, 0, 0, 8, 0},
        {4, 0, 8, 0, 0, 0, 0, 11, 0},
        {0, 8, 0, 7, 0, 4, 0, 0, 2},
        {0, 0, 7, 0, 9, 14, 0, 0, 0},
        {0, 0, 0, 9, 0, 10, 0, 0, 0},
        {0, 0, 4, 14, 10, 0, 2, 0, 0},
        {0, 0, 0, 0, 0, 2, 0, 1, 6},
        {8, 11, 0, 0, 0, 0, 1, 0, 7},
        {0, 0, 2, 0, 0, 0, 6, 7, 0}
    };

    dijkstra(graph, 0);

    return 0;
}

该示例代码中，使用邻接矩阵表示图，图中的顶点编号从0到V-1。在main函数中，我们定义了一个9x9的邻接矩阵图，并调用dijkstra函数求解从顶点0到其它顶点的最短路径长度。最后，调用printSolution函数打印结果。

输出结果如下：