Day60 >> 94、城市间货物运输1️⃣ + 95、城市间货物运输 2️⃣ + 96、城市间货物运输 3️⃣

代码随想录-图论Part10

94、城市间货物运输 1️⃣

Bellman_ford队列优化算法

java 复制代码
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]);
        }
    }
}

95、城市间货物运输 2️⃣

Bellman_ford判断负权回路

java 复制代码
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]);
        }
    }
}

96、城市间货物运输 3️⃣

Bellman_ford单源有限最短路

java 复制代码
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]);
        }
    }
}
相关推荐
2501_933923252 分钟前
Spring Bean作用域揭秘:单例、原型、请求与会话的区别
java·后端·spring·java-ee
wuyk55515 分钟前
【Socket 进阶之路】第 7 章 IO 多路复用 select & poll 完整实战|多 fd 监听、超时等待、优缺点横向对比
服务器·开发语言·网络·数据库·物联网
学编程就要猛20 分钟前
基于Spring AI 的智能聊天机器人
java·spring ai·chat robot
修炼室22 分钟前
Java开发工程师笔试经验贴【高频知识】
java
人工智能培训29 分钟前
大语言模型:从语言理解到通用智能的跃迁
linux·服务器·前端·人工智能
木井巳35 分钟前
【记忆化搜索】最长递增子序列
java·算法·leetcode·深度优先·剪枝·推荐算法
东方护航数据恢复(深圳)38 分钟前
本地小算力设备数据恢复经典案例实操全解_东方护航数据恢复深圳店
大数据·服务器·算法·ai·gpu算力
一拳不是超人40 分钟前
做独立开发三个月,我的第一个产品终于有了 1000 个用户
前端·程序员
杨杨杨大侠42 分钟前
知识库已经有了,Java 程序员还要做什么?Spring AI RAG 实战
java·openai·ai编程
伊信1 小时前
Spring AI 基础学习与应用
java·后端