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

开发技术 开发技术 2022-10-02 次浏览

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 srcdst, 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:

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

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

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