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787. K 站中转内最便宜的航班

有 n 个城市通过一些航班连接。给你一个数组 flights ，其中 flights[i] = [fromi, toi, pricei] ，表示该航班都从城市 fromi 开始，以价格 pricei 抵达 toi。

现在给定所有的城市和航班，以及出发城市 src 和目的地 dst，你的任务是找到出一条最多经过 k 站中转的路线，使得从 src 到 dst 的 价格最便宜 ，并返回该价格。 如果不存在这样的路线，则输出 -1。

 

示例 1：

输入: 
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
输出: 200

解释: 城市航班图如下

从城市 0 到城市 2 在 1 站中转以内的最便宜价格是 200，如图中红色所示。 示例 2：

输入: n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]] src = 0, dst = 2, k = 0 输出: 500 解释: 城市航班图如下

从城市 0 到城市 2 在 0 站中转以内的最便宜价格是 500，如图中蓝色所示。  

提示：

1 <= n <= 100
0 <= flights.length <= (n * (n - 1) / 2)
flights[i].length == 3
0 <= fromi, toi < n
fromi != toi
1 <= pricei <= 104
航班没有重复，且不存在自环
0 <= src, dst, k < n
src != dst

787. Cheapest Flights Within K Stops

There are n cities connected by some number of flights. You are given an array flights where flights[i] = [fromi, toi, pricei] indicates that there is a flight from city fromi to city toi with cost pricei.

You are also given three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. If there is no such route, return -1.

 

Example 1:

Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
Output: 200

Explanation: The graph is shown.

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, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
Output: 500

Explanation: The graph is shown.

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

Constraints:

1 <= n <= 100
0 <= flights.length <= (n * (n - 1) / 2)
flights[i].length == 3
0 <= fromi, toi < n
fromi != toi
1 <= pricei <= 104
There will not be any multiple flights between two cities.
0 <= src, dst, k < n
src != dst

代码

/**
 * @param {number} n
 * @param {number[][]} flights
 * @param {number} src
 * @param {number} dst
 * @param {number} k
 * @return {number}
 */
var findCheapestPrice = function(n, flights, src, dst, k) {
    const INF = 10000 * 101 + 1;
    const f = new Array(k + 2).fill(0).map(() => new Array(n).fill(INF));
    f[0][src] = 0;
    for (let t = 1; t <= k + 1; ++t) {
        for (const flight of flights) {
            const j = flight[0], i = flight[1], cost = flight[2];
            f[t][i] = Math.min(f[t][i], f[t - 1][j] + cost);
        }
    }
    let ans = INF;
    for (let t = 1; t <= k + 1; ++t) {
        ans = Math.min(ans, f[t][dst]);
    }
    return ans == INF ? -1 : ans;
};