README_EN.md

787. Cheapest Flights Within K Stops

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Description

There are n cities connected by m flights. Each flight starts from city u and arrives at v with a price w.

Now given all the cities and flights, 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 graph looks like this:


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 graph looks like this:


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

 

Constraints:

• 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.

Solutions

Python3

from functools import lru_cache


class Solution:
    def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
        @lru_cache(None)
        def dfs(u, k):
            if u == dst:
                return 0
            if k <= 0:
                return float('inf')
            k -= 1
            ans = float('inf')
            for v, p in g[u]:
                ans = min(ans, dfs(v, k) + p)
            return ans

        g = defaultdict(list)
        for u, v, p in flights:
            g[u].append((v, p))
        ans = dfs(src, k + 1)
        return -1 if ans >= float('inf') else ans

Java

class Solution {
    private int[][] memo;
    private int[][] g;
    private int dst;
    private static final int INF = (int) 1e6;

    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        n += 10;
        memo = new int[n][n];
        for (int i = 0; i < n; ++i) {
            Arrays.fill(memo[i], -1);
        }
        g = new int[n][n];
        for (int[] e : flights) {
            g[e[0]][e[1]] = e[2];
        }
        this.dst = dst;
        int ans = dfs(src, k + 1);
        return ans >= INF ? -1 : ans;
    }

    private int dfs(int u, int k) {
        if (memo[u][k] != -1) {
            return memo[u][k];
        }
        if (u == dst) {
            return 0;
        }
        if (k <= 0) {
            return INF;
        }
        int ans = INF;
        for (int v = 0; v < g[u].length; ++v) {
            if (g[u][v] > 0) {
                ans = Math.min(ans, dfs(v, k - 1) + g[u][v]);
            }
        }
        memo[u][k] = ans;
        return ans;
    }
}

C++

class Solution {
public:
    vector<vector<int>> memo;
    vector<vector<int>> g;
    int dst;
    int inf = 1e6;

    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
        n += 10;
        memo.resize(n, vector<int>(n, -1));
        g.resize(n, vector<int>(n));
        for (auto& e : flights) g[e[0]][e[1]] = e[2];
        this->dst = dst;
        int ans = dfs(src, k + 1);
        return ans >= inf ? -1 : ans;
    }

    int dfs(int u, int k) {
        if (memo[u][k] != -1) return memo[u][k];
        if (u == dst) return 0;
        if (k <= 0) return inf;
        int ans = inf;
        for (int v = 0; v < g[u].size(); ++v)
            if (g[u][v] > 0)
                ans = min(ans, dfs(v, k - 1) + g[u][v]);
        memo[u][k] = ans;
        return memo[u][k];
    }
};

Go

func findCheapestPrice(n int, flights [][]int, src int, dst int, k int) int {
	n += 10
	memo := make([][]int, n)
	g := make([][]int, n)
	for i := range memo {
		memo[i] = make([]int, n)
		g[i] = make([]int, n)
		for j := range memo[i] {
			memo[i][j] = -1
		}
	}

	for _, e := range flights {
		g[e[0]][e[1]] = e[2]
	}
	inf := int(1e6)
	var dfs func(u, k int) int
	dfs = func(u, k int) int {
		if memo[u][k] != -1 {
			return memo[u][k]
		}
		if u == dst {
			return 0
		}
		if k <= 0 {
			return inf
		}
		ans := inf
		for v, p := range g[u] {
			if p > 0 {
				ans = min(ans, dfs(v, k-1)+p)
			}
		}
		memo[u][k] = ans
		return ans
	}
	ans := dfs(src, k+1)
	if ans >= inf {
		return -1
	}
	return ans
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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