Algorithm Floyd Warshall最短路径算法错误
问题陈述: 代码:Algorithm Floyd Warshall最短路径算法错误,algorithm,scala,floyd-warshall,Algorithm,Scala,Floyd Warshall,问题陈述: 代码: 导入scala.io.StdIn_ 导入scala.collection.mutable 对象解{ def FloydWarshall(n:Int,adj:Array[Array[Int]]):Array[Array[Int]={ val距离:数组[Array[Int]]=adj 对于(k,程序使用值351作为无效距离的标记,这似乎是问题所在 尽管已知每条边的最大重量为350,但通过由两条或多条边组成的路径,仍然可以达到值为351的距离。程序使用值351作为无效距离的标记,这
导入scala.io.StdIn_
导入scala.collection.mutable
对象解{
def FloydWarshall(n:Int,adj:Array[Array[Int]]):Array[Array[Int]={
val距离:数组[Array[Int]]=adj
对于(k,程序使用值351作为无效距离的标记,这似乎是问题所在
尽管已知每条边的最大重量为350,但通过由两条或多条边组成的路径,仍然可以达到值为351的距离。程序使用值351作为无效距离的标记,这似乎是问题所在
虽然已知每条边的最大权重为350,但是,通过由两条或多条边组成的路径,仍然可以达到值为351的距离。最有可能的问题是,假设使用有效路径无法达到距离351。@qwertyman问题陈述定义了权重r,其中constraint 1虽然350确实是单条边的最大重量,但距离351仍然可以通过由两条或多条边组成的路径到达。@qwertyman是的,这就是问题所在。将此作为答案提交,我将接受。我添加了一个答案,谢谢!最有可能的问题是假设距离351不能通过va到达lid路径。@QWERTIMAN问题陈述定义了权重r,约束条件为1。虽然350确实是单条边的最大权重,但通过由两条或多条边组成的路径仍然可以到达距离351。@QWERTIMAN是的,这就是问题所在。将此作为答案提交,我接受。我添加了一个答案,谢谢!
import scala.io.StdIn._
import scala.collection.mutable
object Solution {
def FloydWarshall(n: Int, adj: Array[Array[Int]]): Array[Array[Int]] = {
val distance: Array[Array[Int]] = adj
for(k <- 0 until n){
for(u <- 0 until n){
for(v <- 0 until n){
distance(u)(v) = minimum(distance(u)(v), distance(u)(k) + distance(k)(v))
}
}
}
distance
}
def minimum(a: Int, b: Int):Int = {
if(a < b){
a
} else {
b
}
}
def main(args: Array[String]) {
var input: Array[Int] = readLine().split(" ").map(_.toInt)
val n: Int = input(0)
val m: Int = input(1)
val adj: Array[Array[Int]] = Array.fill(n, n)(351)
for(_ <- 1 to m){
input = readLine().split(" ").map(_.toInt)
val u: Int = input(0)
val v: Int = input(1)
val w: Int = input(2)
adj(u-1)(v-1) = w
}
for(i <- 0 until n){
adj(i)(i) = 0
}
val q: Int = readInt()
val distance: Array[Array[Int]] = FloydWarshall(n, adj)
val results: mutable.ListBuffer[Int] = mutable.ListBuffer[Int]()
for(_ <- 1 to q) {
input = readLine().split(" ").map(_.toInt)
val u: Int = input(0)
val v: Int = input(1)
val result: Int = if(distance(u-1)(v-1) == 351) -1 else distance(u-1)(v-1)
results += result
}
println(results.mkString("\n"))
}
}
import scala.io.StdIn._
import scala.collection.mutable
import util.control.Breaks._
object Solution {
def FloydWarshall(n: Int, adj: Array[Array[Int]]): Array[Array[Int]] = {
val distance: Array[Array[Int]] = adj
for(k <- 0 until n){
for(u <- 0 until n){
for(v <- 0 until n){
distance(u)(v) = minimum(distance(u)(v), distance(u)(k) + distance(k)(v))
}
}
}
distance
}
def minimum(a: Int, b: Int):Int = {
if(a < b){
a
} else {
b
}
}
def main(args: Array[String]) {
var input: Array[Int] = readLine().split(" ").map(_.toInt)
val n: Int = input(0)
val m: Int = input(1)
val infinity: Int = 350 * 399 + 1// maximum shortest path N-1 edges
val adj: Array[Array[Int]] = Array.fill(n, n)(infinity)
for(_ <- 1 to m){
input = readLine().split(" ").map(_.toInt)
val u: Int = input(0) - 1
val v: Int = input(1) - 1
val w: Int = input(2)
adj(u)(v) = w
}
for(i <- 0 until n){
adj(i)(i) = 0
}
val q: Int = readInt()
val distance: Array[Array[Int]] = FloydWarshall(n, adj)
val results: mutable.ListBuffer[Int] = mutable.ListBuffer[Int]()
for(_ <- 1 to q) {
input = readLine().split(" ").map(_.toInt)
val u: Int = input(0) - 1
val v: Int = input(1) - 1
val result: Int = if(distance(u)(v) == infinity) -1 else distance(u)(v)
results += result
}
println(results.mkString("\n"))
}
}