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网站开发php和ui,除了凡科建站还有什么网站吗,宁夏住房和建设厅网站,一键生成app的软件力扣labuladong一刷day49天迪杰斯特拉 文章目录 力扣labuladong一刷day49天迪杰斯特拉一、743. 网络延迟时间二、1631. 最小体力消耗路径三、1514. 概率最大的路径 一、743. 网络延迟时间 题目链接#xff1a;https://leetcode.cn/problems/network-delay-time/ 使用迪杰斯特…力扣labuladong一刷day49天迪杰斯特拉 文章目录 力扣labuladong一刷day49天迪杰斯特拉一、743. 网络延迟时间二、1631. 最小体力消耗路径三、1514. 概率最大的路径 一、743. 网络延迟时间 题目链接https://leetcode.cn/problems/network-delay-time/ 使用迪杰斯特拉解决加权图某点到所有节点距离的问题。 如果是无权图采用广度优先遍历即可就把图想象成一颗树广度优先就可以说是层序遍历。 而加权图寻找最短距离最经典算法就是迪杰斯特拉算法我们可以计算出某一个点到任意一个之间的最短距离。我们采用优先级队列里面的每一个元素记录的是某点到起点之间的最短距离我们会一直按照最短距离更新这样到达结尾后即可得到最短距离。 class Solution {public int networkDelayTime(int[][] times, int n, int k) {Listint[][] graph new ArrayList[n 1];for (int i 0; i n; i) {graph[i] new ArrayList();}for (int[] time : times) {int a time[0], b time[1], c time[2];graph[a].add(new int[]{b, c});}int[] dist djk(k, graph);int res -1;for (int i 1; i dist.length; i) {if (dist[i] Integer.MAX_VALUE) return -1;res Math.max(res, dist[i]);}return res;}class State {int id;int sToMe;public State(int id, int sToMe) {this.id id;this.sToMe sToMe;}}int[] djk(int start, Listint[][] graph) {int[] dist new int[graph.length];Arrays.fill(dist, Integer.MAX_VALUE);PriorityQueueState pq new PriorityQueue((a, b) - a.sToMe - b.sToMe);dist[start] 0;pq.add(new State(start, 0));while (!pq.isEmpty()) {State poll pq.poll();int id poll.id;int cur poll.sToMe;if (cur dist[id]) continue;for (int[] ints : graph[id]) {int nextId ints[0];int temp dist[id] ints[1];if (temp dist[nextId]) {dist[nextId] temp;pq.add(new State(nextId, temp));}}}return dist;} }二、1631. 最小体力消耗路径 题目链接https://leetcode.cn/problems/path-with-minimum-effort/ 思路基本上就是迪杰斯特拉的典型题目只不过这一次求的是最小消耗但我们在过程中需要求每一条路径的最大消耗在去往下一个点时选择这些最大消耗中的最小消耗做为路径的延伸。 class Solution {public int minimumEffortPath(int[][] heights) {int r heights.length, c heights[0].length;return djk(heights);}Listint[] getList(int x, int y, int r, int c) {int[][] temp {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};Listint[] list new ArrayList();for (int[] ints : temp) {int nx x ints[0], ny y ints[1];if (nx 0 || nx r || ny 0 || ny c) continue;list.add(new int[] {nx, ny});}return list;}class State {int x;int y;int sToMe;public State(int x, int y, int sToMe) {this.x x;this.y y;this.sToMe sToMe;}}int djk(int[][] heights) {int r heights.length, c heights[0].length;int[][] dist new int[r][c];PriorityQueueState pq new PriorityQueue((a, b)- a.sToMe - b.sToMe);for (int i 0; i r; i) {Arrays.fill(dist[i], Integer.MAX_VALUE);}pq.add(new State(0, 0, 0));dist[0][0] 0;while (!pq.isEmpty()) {State poll pq.poll();int x poll.x, y poll.y;int cur poll.sToMe;if (x r-1 y c-1) return cur;if (cur dist[x][y]) continue;for (int[] ints : getList(x, y, r, c)) {int nLen Math.max(dist[x][y], Math.abs(heights[x][y]-heights[ints[0]][ints[1]]));if (nLen dist[ints[0]][ints[1]]) {dist[ints[0]][ints[1]] nLen;pq.add(new State(ints[0], ints[1], nLen));}}}return -1;} }三、1514. 概率最大的路径 题目链接https://leetcode.cn/problems/path-with-maximum-probability/ 思路这题和上一题又有点不一样相当于求最长路径那么需要把优先级队列按照从大到小排序即可。 class Solution {public double maxProbability(int n, int[][] edges, double[] succProb, int start_node, int end_node) {Listdouble[][] graph new ArrayList[n];for (int i 0; i n; i) {graph[i] new ArrayList();}for (int i 0; i edges.length; i) {int from edges[i][0], to edges[i][1];graph[from].add(new double[]{to, succProb[i]});graph[to].add(new double[]{from, succProb[i]});}return djk(graph, start_node, end_node);}class State{int id;double sToE;public State(int id, double sToE) {this.id id;this.sToE sToE;}}double djk(Listdouble[][] graph, int s, int e) {double[] dist new double[graph.length];PriorityQueueState pq new PriorityQueue((a, b)-Double.compare(b.sToE, a.sToE));Arrays.fill(dist, -1);dist[s] 1;pq.add(new State(s, 1));while (!pq.isEmpty()) {State poll pq.poll();int id poll.id;double cur poll.sToE;if (id e) return cur;if (cur dist[id]) continue;for (double[] ints : graph[id]) {int to (int) ints[0];double nLen cur * ints[1];if (nLen dist[to]) {dist[to] nLen;pq.add(new State(to, nLen));}}}return 0.0;} }