给定一个加权无向图G和一个整数S ，任务是打印最短路径的距离以及每个节点从给定顶点S 开始的最短路径数。
例子：
Input: S =1, G =
Output: Shortest Paths distances are : 0 1 2 4 5 3 2 1 3
Numbers of the shortest Paths are: 1 1 1 2 3 1 1 1 2
Explanation:
- The distance of the shortest paths to vertex 1 is 0 and there is only 1 such path, which is {1}.
- The distance of the shortest paths to vertex 2 is 1 and there is only 1 such path, which is {1→2}.
- The distance of the shortest paths to vertex 3 is 2 and there is only 1 such path, which is {1→2→3}.
- The distance of the shortest paths to vertex 4 is 4 and there exist 2 such paths, which are {{1→2→3→4}, {1→2→3→6→4}}.
- The distance of the shortest paths to vertex 5 is 5 and there exist 3 such paths, which are {{1→2→3→4→5}, {1→2→3→6→4→5}, {1→2→3→6→5}}.
- The distance of the shortest paths to vertex 6 is 3 and there is only 1 such path, which is {1→2→3→6}.
- The distance of the shortest paths to vertex 7 is 2 and there is only 1 such path, which is {1→8→7}.
- The distance of the shortest paths to vertex 8 is 1 and there is only 1 such path, which is {1→8}.
- The distance of the shortest paths to vertex 9 is 3 and there exist 2 such paths, which are {{1→8→9}, {1→2→3→9}}.
方法：给定的问题可以使用 Dijkstra 算法解决。请按照以下步骤解决问题：
- 使用 ArrayList
- 初始化两个整数，数组表示Dist[]和Paths[]所有元素都为0以存储每个节点的最短距离以及与源节点S距离最短的路径计数。
- 定义一个函数，比如Dijkstra()来找到每个节点的最短距离并计算距离最短的路径：
- 初始化一个 min PriorityQueue 说PQ和一个字符串的 HashSet 说已解决以存储边缘是否被访问。
- 将0分配给Dist[S] ，将1分配给Paths[S]。
- 现在迭代直到PQ不为空（）并执行以下操作：
- 找到PQ的顶部 Node 并将 Node 值存储在变量u 中。
- 弹出 PQ 的顶部元素。
- 遍历 ArrayList adj[u]并执行以下操作
- 将相邻节点存储在变量 say to 中，将边成本存储在变量 say cost 中：
- 如果访问了边{u, to} ，则继续。
- 如果dist[to]大于dist[u]+cost，则将dist[u]+cost分配给dist[to] ，然后将Paths[u]分配给Paths[to]。
- 否则，如果Paths[to]等于dist[u]+cost，则将Paths[to]增加1。
- 现在，马克，当前边 {u, to}在结算中访问过。
- 调用函数Dijkstra() 。
- 最后，打印数组dist[]和Paths[] 。
下面是上述方法的实现：
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Node class
static class Node implements Comparator {
// Stores the node
public int node;
// Stores the weight
// of the edge
public int cost;
public Node() {}
// Constructor
public Node(int node, int cost)
{
this.node = node;
this.cost = cost;
}
// Costume comparator
@Override
public int compare(Node node1, Node node2)
{
if (node1.cost < node2.cost)
return -1;
if (node1.cost > node2.cost)
return 1;
return 0;
}
}
// Function to insert a node
// in adjacency list
static void addEdge(ArrayList > adj,
int x, int y, int w)
{
adj.get(x).add(new Node(y, w));
adj.get(y).add(new Node(x, w));
}
// Auxiliary function to find shortest paths using
// Dijekstra
static void dijkstra(ArrayList > adj,
int src, int n, int dist[],
int paths[])
{
// Stores the distances of every node in shortest
// order
PriorityQueue pq
= new PriorityQueue(n + 1, new Node());
// Stores if a vertex has been visited or not
Set settled = new HashSet();
// Adds the source node with 0 distance to pq
pq.add(new Node(src, 0));
dist[src] = 0;
paths[src] = 1;
// While pq is not empty()
while (!pq.isEmpty()) {
// Stores the top node of pq
int u = pq.peek().node;
// Stores the distance
// of node u from s
int d = pq.peek().cost;
// Pop the top element
pq.poll();
for (int i = 0; i < adj.get(u).size(); i++) {
int to = adj.get(u).get(i).node;
int cost = adj.get(u).get(i).cost;
// If edge is marked
if (settled.contains(to + " " + u))
continue;
// If dist[to] is greater
// than dist[u] + cost
if (dist[to] > dist[u] + cost) {
// Add the node to to the pq
pq.add(new Node(to, d + cost));
// Update dist[to]
dist[to] = dist[u] + cost;
// Update paths[to]
paths[to] = paths[u];
}
// Otherwise
else if (dist[to] == dist[u] + cost) {
paths[to] = (paths[to] + paths[u]);
}
// Mark the edge visited
settled.add(to + " " + u);
}
}
}
// Function to find the count of shortest path and
// distances from source node to every other node
static void
findShortestPaths(ArrayList > adj,
int s, int n)
{
// Stores the distances of a
// node from source node
int[] dist = new int[n + 5];
// Stores the count of shortest
// paths of a node from
// source node
int[] paths = new int[n + 5];
for (int i = 0; i <= n; i++)
dist[i] = Integer.MAX_VALUE;
for (int i = 0; i <= n; i++)
paths[i] = 0;
// Function call to find
// the shortest paths
dijkstra(adj, s, n, dist, paths);
System.out.print("Shortest Paths distances are : ");
for (int i = 1; i <= n; i++) {
System.out.print(dist[i] + " ");
}
System.out.println();
System.out.print(
"Numbers of the shortest Paths are: ");
for (int i = 1; i <= n; i++)
System.out.print(paths[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
// Input
int N = 9;
int M = 14;
ArrayList > adj = new ArrayList<>();
for (int i = 0; i <= N; i++) {
adj.add(new ArrayList());
}
addEdge(adj, 1, 2, 1);
addEdge(adj, 2, 3, 1);
addEdge(adj, 3, 4, 2);
addEdge(adj, 4, 5, 1);
addEdge(adj, 5, 6, 2);
addEdge(adj, 6, 7, 2);
addEdge(adj, 7, 8, 1);
addEdge(adj, 8, 1, 1);
addEdge(adj, 2, 8, 2);
addEdge(adj, 3, 9, 1);
addEdge(adj, 8, 9, 2);
addEdge(adj, 7, 9, 2);
addEdge(adj, 3, 6, 1);
addEdge(adj, 4, 6, 1);
// Function call
findShortestPaths(adj, 1, N);
}
} 输出：
Shortest Paths distances are : 0 1 2 4 5 3 2 1 3
Numbers of the shortest Paths are: 1 1 1 2 3 1 1 1 2
时间复杂度： O(M + N * log(N))
辅助空间： O(M)
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