Minimum Cost of a Path With Special Roads
<p>You are given an array <code>start</code> where <code>start = [startX, startY]</code> represents your initial position <code>(startX, startY)</code> in a 2D space. You are also given the array <code>target</code> where <code>target = [targetX, targetY]</code> represents your target position <code>(targetX, targetY)</code>.</p>

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

<p>There are also some <strong>special roads</strong>. You are given a 2D array <code>specialRoads</code> where <code>specialRoads[i] = [x1<sub>i</sub>, y1<sub>i</sub>, x2<sub>i</sub>, y2<sub>i</sub>, cost<sub>i</sub>]</code> indicates that the <code>i<sup>th</sup></code> special road goes in <strong>one direction</strong> from <code>(x1<sub>i</sub>, y1<sub>i</sub>)</code> to <code>(x2<sub>i</sub>, y2<sub>i</sub>)</code> with a cost equal to <code>cost<sub>i</sub></code>. You can use each special road any number of times.</p>

<p>Return the <strong>minimum</strong> cost required to go from <code>(startX, startY)</code> to <code>(targetX, targetY)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">5</span></p>

<p><strong>Explanation:</strong></p>

<ol>
	<li>(1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1.</li>
	<li>(1,2) to (3,3). Use <code><span class="example-io">specialRoads[0]</span></code><span class="example-io"> with</span><span class="example-io"> the cost 2.</span></li>
	<li><span class="example-io">(3,3) to (3,4) with a cost of |3 - 3| + |4 - 3| = 1.</span></li>
	<li><span class="example-io">(3,4) to (4,5). Use </span><code><span class="example-io">specialRoads[1]</span></code><span class="example-io"> with the cost</span><span class="example-io"> 1.</span></li>
</ol>

<p><span class="example-io">So the total cost is 1 + 2 + 1 + 1 = 5.</span></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">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]]</span></p>

<p><strong>Output:</strong> <span class="example-io">7</span></p>

<p><strong>Explanation:</strong></p>

<p>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.</p>

<p>Note that the <span class="example-io"><code>specialRoads[0]</code> is directed from (5,7) to (3,2).</span></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">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]]</span></p>

<p><strong>Output:</strong> <span class="example-io">8</span></p>

<p><strong>Explanation:</strong></p>

<ol>
	<li>(1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1.</li>
	<li>(1,2) to (7,4). Use <code><span class="example-io">specialRoads[1]</span></code><span class="example-io"> with the cost</span><span class="example-io"> 4.</span></li>
	<li>(7,4) to (10,4) with a cost of |10 - 7| + |4 - 4| = 3.</li>
</ol>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>start.length == target.length == 2</code></li>
	<li><code>1 &lt;= startX &lt;= targetX &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= startY &lt;= targetY &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= specialRoads.length &lt;= 200</code></li>
	<li><code>specialRoads[i].length == 5</code></li>
	<li><code>startX &lt;= x1<sub>i</sub>, x2<sub>i</sub> &lt;= targetX</code></li>
	<li><code>startY &lt;= y1<sub>i</sub>, y2<sub>i</sub> &lt;= targetY</code></li>
	<li><code>1 &lt;= cost<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
</ul>
Graph; Array; Shortest Path; Heap (Priority Queue)
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 }