There are n cities connected by m flights. Each fight starts from city u and arrives at v with a price w.
Now given all the cities and fights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.
Example 1: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] src = 0, dst = 2, k = 1 Output: 200 Explanation: The graph looks like this: The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture. Example 2: Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] src = 0, dst = 2, k = 0 Output: 500 Explanation: The graph looks like this: The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.Note:
The number of nodes n will be in range [1, 100], with nodes labeled from 0 to n - 1.The size of flights will be in range [0, n * (n - 1) / 2].The format of each flight will be (src, dst, price).The price of each flight will be in the range [1, 10000].k is in the range of [0, n - 1].There will not be any duplicated flights or self cycles. 这道题一看就知道，是求最短路径的题目。所以显然使用 dijkstra 算法，唯一有变化的就是需要对 step 作出限制。
需要注意的是，dijkstra算法中每个循环都能得到一个结点到源的最短路径，但是并不是每个循环 step 都会+1。
package leetcode; import java.util.Arrays; public class Cheapest_Flights_Within_K_Stops_787 { public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) { int[][] matrix = new int[n][n]; for (int i = 0; i < flights.length; i++) { matrix[flights[i][0]][flights[i][1]] = flights[i][2]; } int[] step = new int[n];//记录距离src有step步 int[] distance = new int[n]; Arrays.fill(distance, Integer.MAX_VALUE); int[] visited = new int[n];// 判断是否已确定visited[i]的最短距离 distance[src] = 0; visited[src] = 1; for(int i=1;i<n;i++){ //一次去掉一个node，src已经去掉，还要去掉n-1个node int min = Integer.MAX_VALUE; int minNode = src; for(int j=0;j<n;j++){ if(visited[j]==0&&distance[j]<min){ minNode=j; min=distance[j]; } } visited[minNode]=1; for (int j = 0; j < n; j++) { if (visited[j] == 1) { continue; } if (matrix[minNode][j] != 0 &&distance[minNode] + matrix[minNode][j] < distance[j] &&step[minNode]+1<=K+1) { //因为src到下一个点，也算了一步step，所以要K+1 distance[j] = distance[minNode] + matrix[minNode][j]; step[j]=step[minNode]+1; } } if(minNode==dst){ break; } } if (distance[dst] == Integer.MAX_VALUE) { return -1; } else { return distance[dst]; } } public static void main(String[] args) { // TODO Auto-generated method stub Cheapest_Flights_Within_K_Stops_787 c = new Cheapest_Flights_Within_K_Stops_787(); int[][] flights1 = new int[][] { { 0, 1, 100 }, { 1, 2, 100 }, { 0, 2, 500 } }; System.out.println(c.findCheapestPrice(3, flights1, 0, 2, 1)); int[][] flights2 = new int[][] { { 4, 1, 1 }, { 1, 2, 3 }, { 0, 3, 2 }, { 0, 4, 10 }, { 3, 1, 1 }, { 1, 4, 3 } }; System.out.println(c.findCheapestPrice(5, flights2, 2, 1, 1)); int[][] flights3 = new int[][] { {16,1,81},{15,13,47},{1,0,24},{5,10,21},{7,1,72},{0,4,88}, {16,4,39},{9,3,25},{10,11,28},{13,8,93},{10,3,62},{14,0,38}, {3,10,58},{3,12,46},{3,8,2},{10,16,27},{6,9,90},{14,8,6}, {0,13,31},{6,4,65},{14,17,29},{13,17,64},{12,5,26},{12,1,9}, {12,15,79},{16,11,79},{16,15,17},{4,0,21},{15,10,75},{3,17,23} ,{8,5,55},{9,4,19},{0,10,83},{3,7,17},{0,12,31},{11,5,34}, {17,14,98},{11,14,85},{16,7,48},{12,6,86},{5,17,72}, {4,12,5},{12,10,23},{3,2,31},{12,7,5},{6,13,30},{6,7,88}, {2,17,88},{6,8,98},{0,7,69},{10,15,13},{16,14,24},{1,17,24}, {13,9,82},{13,6,67},{15,11,72},{12,0,83},{1,4,37},{12,9,36}, {9,17,81},{9,15,62},{8,15,71},{10,12,25},{7,6,23},{16,5,76}, {7,17,4},{3,11,82},{2,11,71},{8,4,11},{14,10,51},{8,10,51}, {4,1,57},{6,16,68},{3,9,100},{1,14,26},{10,7,14},{8,17,24}, {1,11,10},{2,9,85},{9,6,49},{11,4,95} }; System.out.println(c.findCheapestPrice(18, flights3, 7, 2, 6)); } }虽然我上面的解法AC了，但是还是有问题的。这其实不是一道最短路径题目。为什么呢？看如下的例子：
测试用例：4, {{0, 1, 1}, {0,2,5},{1,2,1},{2,3,1}}, 0, 3, 1. 正确答案是6，但是我的代码返回-1.
因为单纯的求最短路径不正确。因为有可能源点到目标点是存在路径的，但是最短路径的step超过了限制，此时用上述算法就会得到-1，认为它们之间没有路径，这是不对的。
所以正确的解法应该是用BFS。让我们看看solutions吧。https://leetcode.com/problems/cheapest-flights-within-k-stops/solution/（现在时间2018年2月19日22:45:10，此时solution仍有一些错误，让我们等待后续。）下面这是discuss中的BFS解答。
class Solution { public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) { int[] cost = new int[n]; Arrays.fill(cost, Integer.MAX_VALUE); cost[src] = 0; int[][] adj = new int[n][n]; for(int[] f : flights) { adj[f[0]][f[1]] = f[2]; } Queue<Integer> q = new LinkedList<>(); q.add(src); while(K-- >= 0 && !q.isEmpty()) { int size = q.size(); for(int i = 0; i < size; i++) { int cur = q.poll(); for(int[] f : flights) { int u = f[0], v = f[1]; if(u == cur && cost[v] > cost[u] + adj[u][v]) { cost[v] = cost[u] + adj[u][v]; q.add(v); } } } } return cost[dst] == Integer.MAX_VALUE ? -1 : cost[dst]; } }