[pull] master from youngyangyang04:master #475

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Expand Up		@@ -352,6 +352,77 @@ SPFA(队列优化版Bellman_ford) 在理论上时间复杂度更胜一筹
		## 其他语言版本

		### Java
	```Java
	import java.util.*;

	public class Main {

	// Define an inner class Edge
	static class Edge {
	int from;
	int to;
	int val;
	public Edge(int from, int to, int val) {
	this.from = from;
	this.to = to;
	this.val = val;
	}
	}

	public static void main(String[] args) {
	// Input processing
	Scanner sc = new Scanner(System.in);
	int n = sc.nextInt();
	int m = sc.nextInt();
	List<List<Edge>> graph = new ArrayList<>();

	for (int i = 0; i <= n; i++) {
	graph.add(new ArrayList<>());
	}

	for (int i = 0; i < m; i++) {
	int from = sc.nextInt();
	int to = sc.nextInt();
	int val = sc.nextInt();
	graph.get(from).add(new Edge(from, to, val));
	}

	// Declare the minDist array to record the minimum distance form current node to the original node
	int[] minDist = new int[n + 1];
	Arrays.fill(minDist, Integer.MAX_VALUE);
	minDist[1] = 0;

	// Declare a queue to store the updated nodes instead of traversing all nodes each loop for more efficiency
	Queue<Integer> queue = new LinkedList<>();
	queue.offer(1);

	// Declare a boolean array to record if the current node is in the queue to optimise the processing
	boolean[] isInQueue = new boolean[n + 1];

	while (!queue.isEmpty()) {
	int curNode = queue.poll();
	isInQueue[curNode] = false; // Represents the current node is not in the queue after being polled
	for (Edge edge : graph.get(curNode)) {
	if (minDist[edge.to] > minDist[edge.from] + edge.val) { // Start relaxing the edge
	minDist[edge.to] = minDist[edge.from] + edge.val;
	if (!isInQueue[edge.to]) { // Don't add the node if it's already in the queue
	queue.offer(edge.to);
	isInQueue[edge.to] = true;
	}
	}
	}
	}

	// Outcome printing
	if (minDist[n] == Integer.MAX_VALUE) {
	System.out.println("unconnected");
	} else {
	System.out.println(minDist[n]);
	}
	}
	}

	```

		### Python

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57 changes: 57 additions & 0 deletions problems/kamacoder/0094.城市间货物运输I.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -392,6 +392,63 @@ Bellman_ford 是可以计算负权值的单源最短路算法。
		## 其他语言版本

		### Java
	```Java
	public class Main {

	// Define an inner class Edge
	static class Edge {
	int from;
	int to;
	int val;
	public Edge(int from, int to, int val) {
	this.from = from;
	this.to = to;
	this.val = val;
	}
	}

	public static void main(String[] args) {
	// Input processing
	Scanner sc = new Scanner(System.in);
	int n = sc.nextInt();
	int m = sc.nextInt();
	List<Edge> edges = new ArrayList<>();

	for (int i = 0; i < m; i++) {
	int from = sc.nextInt();
	int to = sc.nextInt();
	int val = sc.nextInt();
	edges.add(new Edge(from, to, val));
	}

	// Represents the minimum distance from the current node to the original node
	int[] minDist = new int[n + 1];

	// Initialize the minDist array
	Arrays.fill(minDist, Integer.MAX_VALUE);
	minDist[1] = 0;

	// Starts the loop to relax all edges n - 1 times to update minDist array
	for (int i = 1; i < n; i++) {

	for (Edge edge : edges) {
	// Updates the minDist array
	if (minDist[edge.from] != Integer.MAX_VALUE && (minDist[edge.from] + edge.val) < minDist[edge.to]) {
	minDist[edge.to] = minDist[edge.from] + edge.val;
	}
	}
	}

	// Outcome printing
	if (minDist[n] == Integer.MAX_VALUE) {
	System.out.println("unconnected");
	} else {
	System.out.println(minDist[n]);
	}
	}
	}

	```

		### Python

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86 changes: 86 additions & 0 deletions problems/kamacoder/0095.城市间货物运输II.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -244,6 +244,92 @@ int main() {
		## 其他语言版本

		### Java
	```Java
	import java.util.*;

	public class Main {
	// 基于Bellman_ford-SPFA方法
	// Define an inner class Edge
	static class Edge {
	int from;
	int to;
	int val;
	public Edge(int from, int to, int val) {
	this.from = from;
	this.to = to;
	this.val = val;
	}
	}

	public static void main(String[] args) {
	// Input processing
	Scanner sc = new Scanner(System.in);
	int n = sc.nextInt();
	int m = sc.nextInt();
	List<List<Edge>> graph = new ArrayList<>();

	for (int i = 0; i <= n; i++) {
	graph.add(new ArrayList<>());
	}

	for (int i = 0; i < m; i++) {
	int from = sc.nextInt();
	int to = sc.nextInt();
	int val = sc.nextInt();
	graph.get(from).add(new Edge(from, to, val));
	}

	// Declare the minDist array to record the minimum distance form current node to the original node
	int[] minDist = new int[n + 1];
	Arrays.fill(minDist, Integer.MAX_VALUE);
	minDist[1] = 0;

	// Declare a queue to store the updated nodes instead of traversing all nodes each loop for more efficiency
	Queue<Integer> queue = new LinkedList<>();
	queue.offer(1);

	// Declare an array to record the times each node has been offered in the queue
	int[] count = new int[n + 1];
	count[1]++;

	// Declare a boolean array to record if the current node is in the queue to optimise the processing
	boolean[] isInQueue = new boolean[n + 1];

	// Declare a boolean value to check if there is a negative weight loop inside the graph
	boolean flag = false;

	while (!queue.isEmpty()) {
	int curNode = queue.poll();
	isInQueue[curNode] = false; // Represents the current node is not in the queue after being polled
	for (Edge edge : graph.get(curNode)) {
	if (minDist[edge.to] > minDist[edge.from] + edge.val) { // Start relaxing the edge
	minDist[edge.to] = minDist[edge.from] + edge.val;
	if (!isInQueue[edge.to]) { // Don't add the node if it's already in the queue
	queue.offer(edge.to);
	count[edge.to]++;
	isInQueue[edge.to] = true;
	}

	if (count[edge.to] == n) { // If some node has been offered in the queue more than n-1 times
	flag = true;
	while (!queue.isEmpty()) queue.poll();
	break;
	}
	}
	}
	}

	if (flag) {
	System.out.println("circle");
	} else if (minDist[n] == Integer.MAX_VALUE) {
	System.out.println("unconnected");
	} else {
	System.out.println(minDist[n]);
	}
	}
	}

	```

		### Python

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65 changes: 65 additions & 0 deletions problems/kamacoder/0096.城市间货物运输III.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -636,6 +636,71 @@ dijkstra 是贪心的思路每一次搜索都只会找距离源点最近的非
		## 其他语言版本

		### Java
	```Java
	import java.util.*;

	public class Main {
	// 基于Bellman_for一般解法解决单源最短路径问题
	// Define an inner class Edge
	static class Edge {
	int from;
	int to;
	int val;
	public Edge(int from, int to, int val) {
	this.from = from;
	this.to = to;
	this.val = val;
	}
	}

	public static void main(String[] args) {
	// Input processing
	Scanner sc = new Scanner(System.in);
	int n = sc.nextInt();
	int m = sc.nextInt();

	List<Edge> graph = new ArrayList<>();

	for (int i = 0; i < m; i++) {
	int from = sc.nextInt();
	int to = sc.nextInt();
	int val = sc.nextInt();
	graph.add(new Edge(from, to, val));
	}

	int src = sc.nextInt();
	int dst = sc.nextInt();
	int k = sc.nextInt();

	int[] minDist = new int[n + 1];
	int[] minDistCopy;

	Arrays.fill(minDist, Integer.MAX_VALUE);
	minDist[src] = 0;

	for (int i = 0; i < k + 1; i++) { // Relax all edges k + 1 times
	minDistCopy = Arrays.copyOf(minDist, n + 1);
	for (Edge edge : graph) {
	int from = edge.from;
	int to = edge.to;
	int val = edge.val;
	// Use minDistCopy to calculate minDist
	if (minDistCopy[from] != Integer.MAX_VALUE && minDist[to] > minDistCopy[from] + val) {
	minDist[to] = minDistCopy[from] + val;
	}
	}
	}

	// Output printing
	if (minDist[dst] == Integer.MAX_VALUE) {
	System.out.println("unreachable");
	} else {
	System.out.println(minDist[dst]);
	}
	}
	}

	```

		### Python

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