2662. Minimum Cost of a Path With Special Roads

Description

You are given an array start where start = [startX, startY] represents your initial position (startX, startY) in a 2D space. You are also given the array target where target = [targetX, targetY] represents your target position (targetX, targetY).

The cost of going from a position (x1, y1) to any other position in the space (x2, y2) is |x2 - x1| + |y2 - y1|.

There are also some special roads. You are given a 2D array specialRoads where specialRoads[i] = [x1_i, y1_i, x2_i, y2_i, cost_i] indicates that the i^th special road goes in one direction from (x1_i, y1_i) to (x2_i, y2_i) with a cost equal to cost_i. You can use each special road any number of times.

Return the minimum cost required to go from (startX, startY) to (targetX, targetY).

Example 1:

Input: start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]

Output: 5

Explanation:

(1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1.
(1,2) to (3,3). Use specialRoads[0] with the cost 2.
(3,3) to (3,4) with a cost of |3 - 3| + |4 - 3| = 1.
(3,4) to (4,5). Use specialRoads[1] with the cost 1.

So the total cost is 1 + 2 + 1 + 1 = 5.

Example 2:

Input: start = [3,2], target = [5,7], specialRoads = [[5,7,3,2,1],[3,2,3,4,4],[3,3,5,5,5],[3,4,5,6,6]]

Output: 7

Explanation:

It is optimal not to use any special edges and go directly from the starting to the ending position with a cost |5 - 3| + |7 - 2| = 7.

Note that the specialRoads[0] is directed from (5,7) to (3,2).

Example 3:

Input: start = [1,1], target = [10,4], specialRoads = [[4,2,1,1,3],[1,2,7,4,4],[10,3,6,1,2],[6,1,1,2,3]]

Output: 8

Explanation:

(1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1.
(1,2) to (7,4). Use specialRoads[1] with the cost 4.
(7,4) to (10,4) with a cost of |10 - 7| + |4 - 4| = 3.

Constraints:

start.length == target.length == 2
1 <= startX <= targetX <= 10⁵
1 <= startY <= targetY <= 10⁵
1 <= specialRoads.length <= 200
specialRoads[i].length == 5
startX <= x1_i, x2_i <= targetX
startY <= y1_i, y2_i <= targetY
1 <= cost_i <= 10⁵

Solutions

Solution 1: Dijkstra

We can find that for each coordinate $(x, y)$ we visit, suppose the minimum cost from the start point to $(x, y)$ is $d$. If we choose to move directly to $(targetX, targetY)$, then the total cost is $d + |x - targetX| + |y - targetY|$. If we choose to go through a special path $(x_1, y_1) \rightarrow (x_2, y_2)$, then we need to spend $|x - x_1| + |y - y_1| + cost$ to move from $(x, y)$ to $(x_2, y_2)$.

Therefore, we can use Dijkstra algorithm to find the minimum cost from the start point to all points, and then choose the smallest one from them.

We define a priority queue $q$, each element in the queue is a triple $(d, x, y)$, which means that the minimum cost from the start point to $(x, y)$ is $d$. Initially, we add $(0, startX, startY)$ to the queue.

In each step, we take out the first element $(d, x, y)$ in the queue, at this time we can update the answer, that is $ans = \min(ans, d + dist(x, y, targetX, targetY))$. Then we enumerate all special paths $(x_1, y_1) \rightarrow (x_2, y_2)$ and add $(d + dist(x, y, x_1, y_1) + cost, x_2, y_2)$ to the queue.

Finally, when the queue is empty, we can get the answer.

The time complexity is $O(n^2 \times \log n)$, and the space complexity is $O(n^2)$. Where $n$ is the number of special paths.

Python3JavaC++GoTypeScript

class Solution:
    def minimumCost(
        self, start: List[int], target: List[int], specialRoads: List[List[int]]
    ) -> int:
        def dist(x1: int, y1: int, x2: int, y2: int) -> int:
            return abs(x1 - x2) + abs(y1 - y2)

        q = [(0, start[0], start[1])]
        vis = set()
        ans = inf
        while q:
            d, x, y = heappop(q)
            if (x, y) in vis:
                continue
            vis.add((x, y))
            ans = min(ans, d + dist(x, y, *target))
            for x1, y1, x2, y2, cost in specialRoads:
                heappush(q, (d + dist(x, y, x1, y1) + cost, x2, y2))
        return ans

class Solution {
    public int minimumCost(int[] start, int[] target, int[][] specialRoads) {
        int ans = 1 << 30;
        int n = 1000000;
        PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        Set<Long> vis = new HashSet<>();
        q.offer(new int[] {0, start[0], start[1]});
        while (!q.isEmpty()) {
            var p = q.poll();
            int x = p[1], y = p[2];
            long k = 1L * x * n + y;
            if (vis.contains(k)) {
                continue;
            }
            vis.add(k);
            int d = p[0];
            ans = Math.min(ans, d + dist(x, y, target[0], target[1]));
            for (var r : specialRoads) {
                int x1 = r[0], y1 = r[1], x2 = r[2], y2 = r[3], cost = r[4];
                q.offer(new int[] {d + dist(x, y, x1, y1) + cost, x2, y2});
            }
        }
        return ans;
    }

    private int dist(int x1, int y1, int x2, int y2) {
        return Math.abs(x1 - x2) + Math.abs(y1 - y2);
    }
}

class Solution {
public:
    int minimumCost(vector<int>& start, vector<int>& target, vector<vector<int>>& specialRoads) {
        auto dist = [](int x1, int y1, int x2, int y2) {
            return abs(x1 - x2) + abs(y1 - y2);
        };
        int ans = 1 << 30;
        int n = 1e6;
        priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> pq;
        pq.push({0, start[0], start[1]});
        unordered_set<long long> vis;
        while (!pq.empty()) {
            auto [d, x, y] = pq.top();
            pq.pop();
            long long k = 1LL * x * n + y;
            if (vis.count(k)) {
                continue;
            }
            vis.insert(k);
            ans = min(ans, d + dist(x, y, target[0], target[1]));
            for (auto& r : specialRoads) {
                int x1 = r[0], y1 = r[1], x2 = r[2], y2 = r[3], cost = r[4];
                pq.push({d + dist(x, y, x1, y1) + cost, x2, y2});
            }
        }
        return ans;
    }
};

func minimumCost(start []int, target []int, specialRoads [][]int) int {
    ans := 1 << 30
    const n int = 1e6
    pq := hp{{0, start[0], start[1]}}
    vis := map[int]bool{}
    for len(pq) > 0 {
        p := pq[0]
        heap.Pop(&pq)
        d, x, y := p.d, p.x, p.y
        if vis[x*n+y] {
            continue
        }
        vis[x*n+y] = true
        ans = min(ans, d+dist(x, y, target[0], target[1]))
        for _, r := range specialRoads {
            x1, y1, x2, y2, cost := r[0], r[1], r[2], r[3], r[4]
            heap.Push(&pq, tuple{d + dist(x, y, x1, y1) + cost, x2, y2})
        }
    }
    return ans
}

func dist(x1, y1, x2, y2 int) int {
    return abs(x1-x2) + abs(y1-y2)
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

type tuple struct {
    d, x, y int
}
type hp []tuple

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

function minimumCost(start: number[], target: number[], specialRoads: number[][]): number {
    const dist = (x1: number, y1: number, x2: number, y2: number): number => {
        return Math.abs(x1 - x2) + Math.abs(y1 - y2);
    };
    const q = new Heap<[number, number, number]>((a, b) => a[0] - b[0]);
    q.push([0, start[0], start[1]]);
    const n = 1000000;
    const vis: Set<number> = new Set();
    let ans = 1 << 30;
    while (q.size()) {
        const [d, x, y] = q.pop();
        const k = x * n + y;
        if (vis.has(k)) {
            continue;
        }
        vis.add(k);
        ans = Math.min(ans, d + dist(x, y, target[0], target[1]));
        for (const [x1, y1, x2, y2, cost] of specialRoads) {
            q.push([d + dist(x, y, x1, y1) + cost, x2, y2]);
        }
    }
    return ans;
}

type Compare<T> = (lhs: T, rhs: T) => number;

class Heap<T = number> {
    data: Array<T | null>;
    lt: (i: number, j: number) => boolean;
    constructor();
    constructor(data: T[]);
    constructor(compare: Compare<T>);
    constructor(data: T[], compare: Compare<T>);
    constructor(data: T[] | Compare<T>, compare?: (lhs: T, rhs: T) => number);
    constructor(
        data: T[] | Compare<T> = [],
        compare: Compare<T> = (lhs: T, rhs: T) => (lhs < rhs ? -1 : lhs > rhs ? 1 : 0),
    ) {
        if (typeof data === 'function') {
            compare = data;
            data = [];
        }
        this.data = [null, ...data];
        this.lt = (i, j) => compare(this.data[i]!, this.data[j]!) < 0;
        for (let i = this.size(); i > 0; i--) this.heapify(i);
    }

    size(): number {
        return this.data.length - 1;
    }

    push(v: T): void {
        this.data.push(v);
        let i = this.size();
        while (i >> 1 !== 0 && this.lt(i, i >> 1)) this.swap(i, (i >>= 1));
    }

    pop(): T {
        this.swap(1, this.size());
        const top = this.data.pop();
        this.heapify(1);
        return top!;
    }

    top(): T {
        return this.data[1]!;
    }
    heapify(i: number): void {
        while (true) {
            let min = i;
            const [l, r, n] = [i * 2, i * 2 + 1, this.data.length];
            if (l < n && this.lt(l, min)) min = l;
            if (r < n && this.lt(r, min)) min = r;
            if (min !== i) {
                this.swap(i, min);
                i = min;
            } else break;
        }
    }

    clear(): void {
        this.data = [null];
    }

    private swap(i: number, j: number): void {
        const d = this.data;
        [d[i], d[j]] = [d[j], d[i]];
    }
}

2662. Minimum Cost of a Path With Special Roads

Description

Solutions

Solution 1: Dijkstra

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