787. Cheapest Flights Within K Stops

787. Cheapest Flights Within K Stops

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 = 1Output: 200

The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the

Example 2:Input: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]src = 0, dst = 2, k = 0Output: 500

The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the

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.

思路： 最短路径的变种，中间限制了结点数。比如当K=1时，表示中间只能进行一次跳转，最长的路径数可以为2。定义：ds[i] 表示从起点src出发，到i，经过最多K次跳转的最短距离。采用松弛法，当经过第k个阶段时，每当有更小的距离时，则更新。

class Solution { public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) { int inf = 0x3f3f3f3f; int[] ds = new int[n]; Arrays.fill(ds, inf); ds[src] = 0; int ans = ds[dst]; for (int k = 0; k <= K; ++k) { int[] nds = new int[n]; Arrays.fill(nds, inf); for (int[] e : flights) { nds[e[1]] = Math.min(nds[e[1]], ds[e[0]] + e[2]); } ds = nds; ans = Math.min(ans, ds[dst]); } if (ans == inf) return -1; else return

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