现在需要你写一个程序来解决最短路径和最便宜路径的问题。这个程序应该能够读入一张城市间高速公路长度和过路费的图，并能根据游客提供的出发地和目的地输出最短路径和最便宜路径。可以使用贪心算法或者Dijkstra算法来实现。用C语言写

好的，我可以为您提供一个基于Dijkstra算法的C语言程序来解决这个问题。

首先，我们需要定义一个图的结构体来存储城市间高速公路长度和过路费的信息：

#define MAX_CITY 100

typedef struct {
    int distance[MAX_CITY][MAX_CITY]; // 城市间距离
    int cost[MAX_CITY][MAX_CITY]; // 城市间过路费
    int num_city; // 城市数量
} Graph;

接下来，我们需要实现Dijkstra算法来计算最短路径和最便宜路径。这里我们使用一个一维数组dist和一个一维数组cost来分别记录从出发点到每个城市的最短路径和最小花费：

#define INF 0x3f3f3f3f

void dijkstra(Graph *graph, int start, int dist[], int cost[]) {
    int visited[MAX_CITY] = {0};
    dist[start] = 0;
    cost[start] = 0;
    for (int i = 0; i < graph->num_city; i++) {
        int min_dist = INF;
        int min_city = -1;
        for (int j = 0; j < graph->num_city; j++) {
            if (!visited[j] && dist[j] < min_dist) {
                min_dist = dist[j];
                min_city = j;
            }
        }
        if (min_city == -1) break;
        visited[min_city] = 1;
        for (int j = 0; j < graph->num_city; j++) {
            if (!visited[j] && graph->distance[min_city][j] != INF) {
                int new_dist = dist[min_city] + graph->distance[min_city][j];
                int new_cost = cost[min_city] + graph->cost[min_city][j];
                if (new_dist < dist[j] || (new_dist == dist[j] && new_cost < cost[j])) {
                    dist[j] = new_dist;
                    cost[j] = new_cost;
                }
            }
        }
    }
}

最后，我们可以利用上述代码来实现一个读入城市间高速公路长度和过路费的图，然后根据游客提供的出发地和目的地输出最短路径和最便宜路径：

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

int main() {
    Graph graph;
    memset(graph.distance, 0x3f, sizeof(graph.distance));
    memset(graph.cost, 0x3f, sizeof(graph.cost));
    graph.num_city = 0;
    int num_road;
    scanf("%d", &num_road);
    for (int i = 0; i < num_road; i++) {
        int city1, city2, distance, cost;
        scanf("%d %d %d %d", &city1, &city2, &distance, &cost);
        graph.distance[city1][city2] = graph.distance[city2][city1] = distance;
        graph.cost[city1][city2] = graph.cost[city2][city1] = cost;
        graph.num_city = (city1 > graph.num_city) ? city1 : graph.num_city;
        graph.num_city = (city2 > graph.num_city) ? city2 : graph.num_city;
    }
    int start, end;
    scanf("%d %d", &start, &end);
    int dist[MAX_CITY], cost[MAX_CITY];
    memset(dist, 0x3f, sizeof(dist));
    memset(cost, 0x3f, sizeof(cost));
    dijkstra(&graph, start, dist, cost);
    printf("最短路径长度：%d，最小花费：%d\n", dist[end], cost[end]);
    return 0;
}

请注意，上述代码只是一个简单的示例程序，实际应用中还需要考虑更多的细节问题，例如输入的城市编号是否合法等等。