你将如何使用二进制堆来实现Dijkstra的算法？我的目标是运行O（M log N）。
假设一个王国中有N个城市，这个王国有M个火车路线，S是首都。
输入是N M S，接着是M分离三联体（U，V和D）的列表，这意味着从U市到V市的列车路线需要D天。请注意，此列车路线只能从城市U到V，而不是从V到U.
输出是一行，包含一个以空格分隔的N个整数列表，其中第I个整数是从城市I到城市S的最小天数。如果从城市I到城市旅行是不可能的S，输出 - 1为第I个整数。
如果样本输入是这样的：
4 4 4
1 4 1
3 1 2
3 4 4
4 2 1
然后输出是：
1 -1 3 0
这是另一个例子：
5 8 2
3 2 2
2 3 2
2 5 2
5 2 2
4 2 2
2 4 2
1 4 2
2 1 2
输出结果为：
4 0 2 2 2
我的目标是尝试使用二进制堆来解决这个问题，但我遇到了麻烦。我现在正在使用邻接列表，我会看看是否可以在此发布代码，但如果你可以帮助我，那将会非常有用。
感谢您的帮助。
编辑：这是我使用邻接列表的代码。
//import static jdk.nashorn.internal.runtime.regexp.joni.Syntax.Java;
import java.util.Scanner;
公共课Dijkstra { public static void main（String [] args）{
int N, M, S;
Scanner scan = new Scanner(System.in);
N = scan.nextInt(); // # cities
M = scan.nextInt(); // # train routes
S = scan.nextInt(); // capital city
// System.out.println(N + " " + M + " " + S);
// NOW THE ARRAYS
int [][] A = new int[50010][60]; // the neighbors of each city
int [][] W = new int[50010][60]; // the weights of going to neighbors
int [] deg = new int[50010]; // the degree of the city
// The limits are 50,010 and 60 because the problem statement said that there are at most
// 50,000 cities, and we just added 10 just to be sure. We have 60 because the maximum number of
// train routes is 50, and we just added 10 to that.
// with each incoming vertex/city, we will at first initialize the degree to be 0
for(int i = 1; i <=N; ++i) {
deg[i] = 0; // initialize the degree of each vertex to 0
}
// this is for each of the train routes
for(int i = 1; i <= M; ++i) {
int u, v, w;
u = scan.nextInt(); // origin
v = scan.nextInt(); // destination
w = scan.nextInt(); // # days
// System.out.println(u + " " + v + " " + w);
// WITH THE ARRAYS
A[u][deg[u]] = v; // adding an edge (u,v) to the graph where u is origin and deg[u] is weight
W[u][deg[u]] = w; // set its weight to w, the number of days it takes
deg[u]++; // increase degree of vertex u by 1
}
//for(int i = 1; i <= N; ++i) {
// System.out.println("vertex:" + i + "'s neighbors");
// for(int j = 0; j < deg[i]; ++j) {
// System.out.println(A[i][j] + " " + W[i][j]);
// }
//}
// compute distance from U (origin) to S (capital city) by Dijkstra's algorithm
// Dijkstra's algorithm: find the shortest path distance from each vertex to the capital
for(int U = 1; U <= N; ++U) {
// INITIALIZATION
int[] visited = new int[50010]; // create an empty array w/ max # cities space for cities that are visited
int[] dist = new int[50010]; // create an empty array w/ max # cities space for distance of each city
// loop that goes through the arrays and fills in values up to N number of cities
for(int V = 1; V <= N; ++V) {
dist[V] = 100000000; // set the distance of the city to the capital to be the maximum possible number
visited[V] = 0; // set the cities that are visited to be 0
}
// ACTUAL ALGORITHM
dist[U] = 0; // set the distance of the city to be 0
for(int k = 1; k <= N; ++k) {
//find an unvisited vertex with minimum distance
int min = 100000000;
int minVertex = 1;
for(int i = 1; i<=N; ++i) {
// if the city has not been visited and the distance from it to the capital is less than the minimum
if(visited[i] == 0 && dist[i] < min) {
min = dist[i]; // set the new minimum to be this distance
minVertex = i; // set the minimum vertex to be this number
}
}
visited[minVertex] = 1; // set this value to 1 to show that the city has been visited
// relax the edges that are adjacent to minVertex to update the shortest path distance to
// neighbors of minVertex
for(int j = 0; j < deg[minVertex]; ++j) { // this is updating the minimum weight of the city
// A[minVertex][j] is the j-th neighbor of minVertex
// W[minVertex][j] is the weight of the corresponding edge
int newDist = dist[minVertex] + W[minVertex][j];
if (newDist < dist[A[minVertex][j]]) {
dist[A[minVertex][j]] = newDist;
}
}
}
if(dist[S] == 100000000) { // if the distance of this city is still the maximum, it does not have a connection
System.out.print("-1 ");
}
else { // if it has a distance less than max, it means there is a minimum distance and we will print that
System.out.print(dist[S] + " ");
}
}
System.out.println("");
}
}