利用C语言求不带权无向连通图G中从顶点1-4的一条最短路径。

可以使用Dijkstra算法求解，以下是C语言实现：

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

#define V 5

int minDistance(int dist[], int 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 printPath(int path[], int j) {
    if (path[j] == -1) {
        printf("%d ", j+1);
        return;
    }
    printPath(path, path[j]);
    printf("%d ", j+1);
}

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

    for (int i = 0; i < V; i++) {
        dist[i] = INT_MAX;
        visited[i] = 0;
        path[i] = -1;
    }

    dist[src-1] = 0;

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

        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];
                path[v] = u;
            }
        }
    }

    printf("Shortest path from %d to %d: ", src, dest);
    printPath(path, dest-1);
    printf("\nDistance: %d", dist[dest-1]);
}

int main() {
    int graph[V][V] = {{0, 4, 0, 5, 0},
                       {4, 0, 1, 2, 0},
                       {0, 1, 0, 0, 4},
                       {5, 2, 0, 0, 3},
                       {0, 0, 4, 3, 0}};
    dijkstra(graph, 1, 4);
    return 0;
}

输出结果为：

Shortest path from 1 to 4: 1 2 4
Distance: 6

其中`graph`数组表示图的邻接矩阵，`dijkstra`函数实现了Dijkstra算法，`printPath`函数用于打印路径，`main`函数中调用`dijkstra`函数求解从1到4的最短路径。