2024-03-18 solved

SeongHoonC/dijkstra/특정한 최단 경로.kt

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+import java.io.BufferedReader
+import java.io.InputStreamReader
+import java.util.*
+import kotlin.math.min
+
+private const val INF = 987_654_321L
+fun main() {
+    val br = BufferedReader(InputStreamReader(System.`in`))
+
+    val (N, E) = br.readLine().split(" ").map { it.toInt() }
+    val graph = Array(N + 1) { mutableListOf<Pair<Int, Long>>() }
+    for (i in 0 until E) {
+        val (start, end, cost) = br.readLine().split(" ").map { it.toInt() }
+        graph[start].add((end to cost.toLong()))
+        graph[end].add((start to cost.toLong()))
+    }
+
+    val (v1, v2) = br.readLine().split(" ").map { it.toInt() }
+    // 1 to v1 + v1 to v2 + v2 to N
+    val cost1 = dijkstra(N, 1, v1, graph) + dijkstra(N, v1, v2, graph) + dijkstra(N, v2, N, graph)
+    // 1 to v2 + v2 to v1 + v1 to N
+    val cost2 = dijkstra(N, 1, v2, graph) + dijkstra(N, v2, v1, graph) + dijkstra(N, v1, N, graph)
+
+    val answer = min(cost1, cost2)
+    println(if (answer >= INF) -1 else answer)
+}
+
+// start to end 최단거리 구하기
+private fun dijkstra(N: Int, start: Int, end: Int, graph: Array<MutableList<Pair<Int, Long>>>): Long {
+    if (start == end) {
+        return 0
+    }
+    val pq = PriorityQueue<Edge>()
+    val distance = Array(N + 1) { INF }
+
+    distance[start] = 0
+    pq.add(Edge(start, 0))
+
+    while (pq.isNotEmpty()) {
+        val (now, dist) = pq.poll()
+        for ((next, cost) in graph[now]) {
+            val nextCost: Long = dist + cost
+            if (distance[next] > nextCost) {
+                distance[next] = nextCost
+                pq.add(Edge(next, nextCost))
+            }
+        }
+    }
+
+    return distance[end]
+}
+
+data class Edge(val start: Int, val cost: Long) : Comparable<Edge> {
+
+    // 거리(비용)가 짧은 것이 높은 우선순위를 가지도록 설정
+    override operator fun compareTo(other: Edge): Int {
+        return if (this.cost < other.cost) -1 else 1
+    }
+}