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社区首页 >专栏 >LWC 72: 787. Cheapest Flights Within K Stops

LWC 72: 787. Cheapest Flights Within K Stops

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发布2019-05-26 00:41:12
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发布2019-05-26 00:41:12
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文章被收录于专栏:机器学习入门机器学习入门

LWC 72: 787. Cheapest Flights Within K Stops

Problem:

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 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 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.

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。

Java版本:

代码语言:javascript
复制
    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 ans;
    }

Python版本:

代码语言:javascript
复制
class Solution(object):
    def findCheapestPrice(self, n, flights, src, dst, K):
        """
        :type n: int
        :type flights: List[List[int]]
        :type src: int
        :type dst: int
        :type K: int
        :rtype: int
        """
        ds = [float('inf')] * n
        ds[src] = 0

        ans = ds[dst]
        for k in range(0, K + 1):
            nds = [float('inf')] * n
            for e in flights:
                nds[e[1]] = min(nds[e[1]], ds[e[0]] + e[2])
            ds = nds
            ans = min(ans, ds[dst])

        return -1 if ans == float('inf') else ans
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原始发表:2018年02月18日,如有侵权请联系 cloudcommunity@tencent.com 删除

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