Leetcode787

問從s到t最多k+1條邊的最小值，不存在返回-1

 

圖論題，不會做，感覺比較難理解，應該算是dijkstra算法的應用

講解dijkstra算法：

https://www.luogu.com.cn/blog/ztyluogucpp/qian-tan-dijkstra

 

我來解釋一下高票裏的解法

維護一個優先隊列，節點內容是到達某個節點用了stop條邊的最小值，優先隊列按照邊權升序排列。

初始化的時候加入一個起點，每次從有限隊列取出一個，然後枚舉從這個點開始的所有出邊，維護後加入到優先隊列，感覺有點說不清楚，具體看代碼吧

 

    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) {
        //建圖
        List<List<Edge>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) graph.add(new ArrayList<>());
        for (int[] a : flights) graph.get(a[0]).add(new Edge(a[2], a[1]));

        //優先隊列初始化, 按照邊權從小到大排序
        PriorityQueue<Node> pq = new PriorityQueue<>(Comparator.comparingInt(o -> o.dis));
        //最多k+1條邊
        pq.add(new Node(src, 0, K + 1));

        while (!pq.isEmpty()) {
            Node poll = pq.poll();
            int dis = poll.dis;
            int point = poll.startPoint;
            int stop = poll.stop;
            //優先隊列裏維護着到達point的最小值
            if (point == dst) return dis;
            if (stop == 0) continue;
            //枚舉從point所有的出邊
            List<Edge> edges = graph.get(point);
            for (Edge edge : edges) {
                pq.add(new Node(edge.endPoint, dis + edge.dis, stop - 1));

            }
        }
        return -1;

    }

    class Edge {
        int dis;
        int endPoint;

        public Edge(int dis, int endPoint) {
            this.dis = dis;
            this.endPoint = endPoint;
        }
    }

    class Node {
        int startPoint;
        int dis;
        int stop;

        public Node(int startPoint, int dis, int stop) {
            this.startPoint = startPoint;
            this.dis = dis;
            this.stop = stop;
        }
    }