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3778. Minimum Distance Excluding One Maximum Weighted Edge πŸ”’

Description

You are given a positive integer n and a 2D integer array edges, where edges[i] = [ui, vi, wi].

There is a weighted connected simple undirected graph with n nodes labeled from 0 to n - 1. Each [ui, vi, wi] in edges represents an edge between node ui and node vi with positive weight wi.

The cost of a path is the sum of weights of the edges in the path, excluding the edge with the maximum weight. If there are multiple edges in the path with the maximum weight, only the first such edge is excluded.

Return an integer representing the minimum cost of a path going from node 0 to node n - 1.

 

Example 1:

Input: n = 5, edges = [[0,1,2],[1,2,7],[2,3,7],[3,4,4]]

Output: 13

Explanation:

There is only one path going from node 0 to node 4: 0 -> 1 -> 2 -> 3 -> 4.

The edge weights on this path are 2, 7, 7, and 4.

Excluding the first edge with maximum weight, which is 1 -> 2, the cost of this path is 2 + 7 + 4 = 13.

Example 2:

Input: n = 3, edges = [[0,1,1],[1,2,1],[0,2,50000]]

Output: 0

Explanation:

There are two paths going from node 0 to node 2:

  • 0 -> 1 -> 2

The edge weights on this path are 1 and 1.

Excluding the first edge with maximum weight, which is 0 -> 1, the cost of this path is 1.

  • 0 -> 2

The only edge weight on this path is 1.

Excluding the first edge with maximum weight, which is 0 -> 2, the cost of this path is 0.

The minimum cost is min(1, 0) = 0.

 

Constraints:

  • 2 <= n <= 5 * 104
  • n - 1 <= edges.length <= 109
  • edges[i] = [ui, vi, wi]
  • 0 <= ui < vi < n
  • [ui, vi] != [uj, vj]
  • 1 <= wi <= 5 * 104
  • The graph is connected.

Solutions

Solution 1

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class Solution:
    def minCostExcludingMax(self, n: int, edges: List[List[int]]) -> int:
        g = [[] for _ in range(n)]
        for u, v, w in edges:
            g[u].append((v, w))
            g[v].append((u, w))
        dist = [[inf] * 2 for _ in range(n)]
        dist[0][0] = 0
        pq = [(0, 0, 0)]
        while pq:
            cur, u, used = heappop(pq)
            if cur > dist[u][used]:
                continue
            if u == n - 1 and used:
                return cur
            for v, w in g[u]:
                nxt = cur + w
                if nxt < dist[v][used]:
                    dist[v][used] = nxt
                    heappush(pq, (nxt, v, used))
                if used == 0:
                    nxt = cur
                    if nxt < dist[v][1]:
                        dist[v][1] = nxt
                        heappush(pq, (nxt, v, 1))
        return dist[n - 1][1]

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class Solution {
    public long minCostExcludingMax(int n, int[][] edges) {
        List<int[]>[] g = new ArrayList[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int u = e[0], v = e[1], w = e[2];
            g[u].add(new int[]{v, w});
            g[v].add(new int[]{u, w});
        }

        long inf = Long.MAX_VALUE / 4;
        long[][] dist = new long[n][2];
        for (int i = 0; i < n; i++) {
            dist[i][0] = inf;
            dist[i][1] = inf;
        }
        dist[0][0] = 0;

        PriorityQueue<long[]> pq = new PriorityQueue<>(Comparator.comparingLong(a -> a[0]));
        pq.add(new long[]{0, 0, 0});

        while (!pq.isEmpty()) {
            long[] t = pq.poll();
            long cur = t[0];
            int u = (int) t[1];
            int used = (int) t[2];

            if (cur > dist[u][used]) {
                continue;
            }
            if (u == n - 1 && used == 1) {
                return cur;
            }

            for (int[] ed : g[u]) {
                int v = ed[0], w = ed[1];
                long nxt = cur + w;
                if (nxt < dist[v][used]) {
                    dist[v][used] = nxt;
                    pq.add(new long[]{nxt, v, used});
                }

                if (used == 0) {
                    nxt = cur;
                    if (nxt < dist[v][1]) {
                        dist[v][1] = nxt;
                        pq.add(new long[]{nxt, v, 1});
                    }
                }
            }
        }

        return dist[n - 1][1];
    }
}

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class Solution {
public:
    long long minCostExcludingMax(int n, vector<vector<int>>& edges) {
        vector<vector<pair<int, int>>> g(n);
        for (auto& e : edges) {
            int u = e[0], v = e[1], w = e[2];
            g[u].push_back({v, w});
            g[v].push_back({u, w});
        }

        long long inf = LLONG_MAX / 4;
        vector<array<long long, 2>> dist(n, {inf, inf});
        dist[0][0] = 0;

        priority_queue<array<long long, 3>, vector<array<long long, 3>>, greater<array<long long, 3>>> pq;
        pq.push({0, 0, 0});

        while (!pq.empty()) {
            auto t = pq.top();
            pq.pop();
            long long cur = t[0];
            int u = t[1];
            int used = t[2];

            if (cur > dist[u][used]) {
                continue;
            }
            if (u == n - 1 && used == 1) {
                return cur;
            }

            for (auto [v, w] : g[u]) {
                long long nxt = cur + w;
                if (nxt < dist[v][used]) {
                    dist[v][used] = nxt;
                    pq.push({nxt, v, used});
                }

                if (used == 0) {
                    nxt = cur;
                    if (nxt < dist[v][1]) {
                        dist[v][1] = nxt;
                        pq.push({nxt, v, 1});
                    }
                }
            }
        }

        return dist[n - 1][1];
    }
};

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func minCostExcludingMax(n int, edges [][]int) int64 {
    g := make([][]edge, n)
    for _, e := range edges {
        u, v, w := e[0], e[1], e[2]
        g[u] = append(g[u], edge{v, w})
        g[v] = append(g[v], edge{u, w})
    }

    inf := int64(math.MaxInt64 / 4)
    dist := make([][2]int64, n)
    for i := 0; i < n; i++ {
        dist[i][0] = inf
        dist[i][1] = inf
    }
    dist[0][0] = 0

    pq := hp{{0, 0, 0}}
    for len(pq) > 0 {
        t := heap.Pop(&pq).(state)
        cur, u, used := t.cur, t.u, t.used
        if cur > dist[u][used] {
            continue
        }
        if u == n-1 && used == 1 {
            return cur
        }
        for _, ed := range g[u] {
            v, w := ed.to, int64(ed.w)

            nxt := cur + w
            if nxt < dist[v][used] {
                dist[v][used] = nxt
                heap.Push(&pq, state{nxt, v, used})
            }

            if used == 0 {
                nxt = cur
                if nxt < dist[v][1] {
                    dist[v][1] = nxt
                    heap.Push(&pq, state{nxt, v, 1})
                }
            }
        }
    }
    return dist[n-1][1]
}

type edge struct {
    to int
    w  int
}

type state struct {
    cur  int64
    u    int
    used int
}

type hp []state

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].cur < h[j].cur }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(x any)        { *h = append(*h, x.(state)) }
func (h *hp) Pop() (x any)      { a := *h; x = a[len(a)-1]; *h = a[:len(a)-1]; return }

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function minCostExcludingMax(n: number, edges: number[][]): number {
    const g: [number, number][][] = Array.from({ length: n }, () => []);
    for (const [u, v, w] of edges) {
        g[u].push([v, w]);
        g[v].push([u, w]);
    }

    const INF = Infinity;
    const dist: number[][] = Array.from({ length: n }, () => [INF, INF]);
    dist[0][0] = 0;

    const pq = new PriorityQueue<[number, number, number]>((a, b) =>
        a[0] === b[0] ? a[1] - b[1] : a[0] - b[0],
    );

    pq.enqueue([0, 0, 0]);

    while (pq.size() > 0) {
        const [cur, u, used] = pq.dequeue()!;
        if (cur > dist[u][used]) {
            continue;
        }
        if (u === n - 1 && used === 1) {
            return cur;
        }

        for (const [v, w] of g[u]) {
            const nxt1 = cur + w;
            if (nxt1 < dist[v][used]) {
                dist[v][used] = nxt1;
                pq.enqueue([nxt1, v, used]);
            }
            if (used === 0) {
                const nxt2 = cur;
                if (nxt2 < dist[v][1]) {
                    dist[v][1] = nxt2;
                    pq.enqueue([nxt2, v, 1]);
                }
            }
        }
    }

    return dist[n - 1][1];
}

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