787. Cheapest Flights Within K Stops 无坐标图，能停k站，最便宜的飞机

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 = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
Output: 700
Explanation:
The graph is shown above.
The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.
Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.

Example 2:

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 above.
The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.

Example 3:

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 above.
The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.

 

不知道怎么判断这样的连通性：
map.get(i[0]).add(new int[]{i[1],i[2]});至少可以这样 用个map存
然后用if(k<0)来退出dfs。
感觉不是特别普遍啊，算了，这道题就这样吧。

 

class Solution {
    int ans_dfs;
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K)
    {
        ans_dfs=Integer.MAX_VALUE;
        Map<Integer,List<int[]>> map=new HashMap<>();
        for(int[] i:flights)
        {
            map.putIfAbsent(i[0],new ArrayList<>());
            map.get(i[0]).add(new int[]{i[1],i[2]});
        }
        dfs(map,src,dst,K+1,0);
        return ans_dfs==Integer.MAX_VALUE?-1:ans_dfs;
    }
    public void dfs(Map<Integer,List<int[]>> map, int src, int dst, int k, int cost)
    {
        if(k<0)
            return;
        if(src==dst)
        {
            ans_dfs=cost;
            return;
        }
        if(!map.containsKey(src))
            return;
        for(int[] i:map.get(src))
        {
            if(cost+i[1]>ans_dfs)               //Pruning, check the sum of current price and next cost. If it's greater then the ans so far, continue
                continue;
            dfs(map,i[0],dst,k-1,cost+i[1]);
        }
    }
}