基本上,使用 Floyd-Warshall 算法的目的是确定连通图中两个节点之间的最短路径。我想做的不是简单地找到最短路径,而是想要权重均匀的最短路径。
例如,这是 Floyd-Warshall 算法的简单实现:
#include <stdio.h>
main()
{
int dist[10][10],succ[10][10],n,i,j,k;
int newDist;
scanf("%d",&n);
for (i=0;i<n;i++)
for (j=0;j<n;j++)
{
dist[i][j]=999;
succ[i][j]=j;
}
while (1)
{
scanf("%d %d %d",&i,&j,&k);
if (i==(-1))
break;
dist[i][j]=k;
distOdd[i][j]=k;
distEven[i][j]=k;
}
printf(" ");
for (i=0;i<n;i++)
printf("%3d ",i);
printf("\n");
for (i=0;i<n;i++)
{
printf("%3d ",i);
for (k=0;k<n;k++)
printf("%3d %d ",dist[i][k],succ[i][k]);
printf("\n");
}
printf("-------------------------------\n");
/* Floyd-Warshall */
for (j=0;j<n;j++)
{
for (i=0;i<n;i++)
if (dist[i][j]<999)
for (k=0;k<n;k++)
{
newDist=dist[i][j]+dist[j][k];
if (newDist<dist[i][k])
{
dist[i][k]=newDist;
succ[i][k]=succ[i][j];
}
}
printf(" ");
for (i=0;i<n;i++)
printf("%3d ",i);
printf("\n");
for (i=0;i<n;i++)
{
printf("%3d ",i);
for (k=0;k<n;k++)
printf("%3d %d ",dist[i][k],succ[i][k]);
printf("\n");
}
printf("-------------------------------\n");
}
for (i=0;i<n;i++)
for (j=0;j<n;j++)
if (dist[i][j]==999)
printf("No path from %d to %d\n",i,j);
else
{
printf("Distance %d for %d ",dist[i][j],i);
for (k=succ[i][j];
k!=j;
k=succ[k][j])
printf("%d ",k);
printf("%d\n",j);
}
}
给定以下输入:
6
0 1 1
1 2 1
2 3 1
3 1 1
1 4 1
4 5 1
-1 -1 -1
我想要以下输出(忽略格式,我只需要一种方法来找到“每一步的奇数矩阵”
initial odd matrix
999 0 1 1 999 2 999 3 999 4 999 5
999 0 999 1 1 2 999 3 1 4 999 5
999 0 999 1 999 2 1 3 999 4 999 5
999 0 1 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 1 5
999 0 999 1 999 2 999 3 999 4 999 5
-------------------------------
Process column 0
odd matrix
999 0 1 1 999 2 999 3 999 4 999 5
999 0 999 1 1 2 999 3 1 4 999 5
999 0 999 1 999 2 1 3 999 4 999 5
999 0 1 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 1 5
999 0 999 1 999 2 999 3 999 4 999 5
even matrix
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
-------------------------------
Process column 1
odd matrix
999 0 1 1 999 2 999 3 999 4 999 5
999 0 999 1 1 2 999 3 1 4 999 5
999 0 999 1 999 2 1 3 999 4 999 5
999 0 1 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 1 5
999 0 999 1 999 2 999 3 999 4 999 5
even matrix
999 0 999 1 2 1 999 3 2 1 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 2 1 999 3 2 1 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
-------------------------------
Process column 2
odd matrix
999 0 1 1 999 2 3 1 999 4 999 5
999 0 999 1 1 2 999 3 1 4 999 5
999 0 999 1 999 2 1 3 999 4 999 5
999 0 1 1 999 2 3 1 999 4 999 5
999 0 999 1 999 2 999 3 999 4 1 5
999 0 999 1 999 2 999 3 999 4 999 5
even matrix
999 0 999 1 2 1 999 3 2 1 999 5
999 0 999 1 999 2 2 2 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 2 1 999 3 2 1 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
-------------------------------
Process column 3
odd matrix
999 0 1 1 5 1 3 1 5 1 999 5
999 0 3 2 1 2 5 2 1 4 999 5
999 0 5 3 3 3 1 3 3 3 999 5
999 0 1 1 5 1 3 1 5 1 999 5
999 0 999 1 999 2 999 3 999 4 1 5
999 0 999 1 999 2 999 3 999 4 999 5
even matrix
999 0 4 1 2 1 6 1 2 1 999 5
999 0 6 2 4 2 2 2 4 2 999 5
999 0 2 3 6 3 4 3 6 3 999 5
999 0 4 1 2 1 6 1 2 1 999 5
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
-------------------------------
Process column 4
odd matrix
999 0 1 1 5 1 3 1 5 1 3 1
999 0 3 2 1 2 5 2 1 4 5 2
999 0 5 3 3 3 1 3 3 3 7 3
999 0 1 1 5 1 3 1 5 1 3 1
999 0 999 1 999 2 999 3 999 4 1 5
999 0 999 1 999 2 999 3 999 4 999 5
even matrix
999 0 4 1 2 1 6 1 2 1 6 1
999 0 6 2 4 2 2 2 4 2 2 4
999 0 2 3 6 3 4 3 6 3 4 3
999 0 4 1 2 1 6 1 2 1 6 1
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
-------------------------------
Process column 5
odd matrix
999 0 1 1 5 1 3 1 5 1 3 1
999 0 3 2 1 2 5 2 1 4 5 2
999 0 5 3 3 3 1 3 3 3 7 3
999 0 1 1 5 1 3 1 5 1 3 1
999 0 999 1 999 2 999 3 999 4 1 5
999 0 999 1 999 2 999 3 999 4 999 5
even matrix
999 0 4 1 2 1 6 1 2 1 6 1
999 0 6 2 4 2 2 2 4 2 2 4
999 0 2 3 6 3 4 3 6 3 4 3
999 0 4 1 2 1 6 1 2 1 6 1
999 0 999 1 999 2 999 3 999 4 999 5
999 0 999 1 999 2 999 3 999 4 999 5
-------------------------------
我的代码当前所做的是获得最佳权重,该权重在每个单独的“奇数”和“偶数”矩阵中表示。
我缺乏理解的是,当最佳值位于相反矩阵(奇/偶)时,“奇”和“偶”矩阵如何得出非最佳值。在我看来,必须有某种循环或递归才能做到这一点,但我不知道如何实现这一点。
最佳答案
不是用 C 语言,但这应该不是问题。我相信 F-W 需要进行两次修改才能获得最短的奇数/偶数路径:
它需要运行两次。这是因为如果路径自身循环,则可能会改变均匀度。如下图所示:A --5--> B --2-->(回到 A)。为了沿着均匀的路径从 A 到 B,我们需要走 A-B-A-B。但是,如果我们不能通过运行两次获得一定均匀度的路径,那么运行两次以上就没有意义了。
您需要尝试所有组合才能找到更好的路径(请参阅从 0 到 1 的最内层循环)。通过添加奇数边缘等,迄今为止的偶数路径可能会成为新的最佳奇数路径。
我认为这个算法应该是正确的,如果你发现任何错误,请随时骂我。 >D
编辑:添加路径内存(用//ADDED 标记的部分)。当然,这使得算法内存效率低下,因此只有在确实需要时才应使用。 ATM 我想不出一种方法可以让标准 F-W 路径重建在这种情况下发挥作用。由于路径可能比顶点数量长,我不确定路径重建将如何工作。我还没有广泛测试路径内存,所以它可能会被窃听。不过可能工作正常。
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace ConsoleApplication5
{
class Program
{
static void Main(string[] args)
{
int n = 5;
// Generate graph
Random r = new Random(1);
// ADDED
List<int>[,,] path = new List<int>[n, n, 2];
int[,,] cost = new int[n, n, 2];
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// ADDED
path[i, j, 0] = new List<int>{i};
path[i, j, 1] = new List<int>{i};
if (i == j)
{
cost[i, j, 0] = 0;
cost[i, j, 1] = -1;
continue;
}
int x = r.Next() % 9 + 1;
if (r.Next(100) < 60)
{
cost[i, j, 0] = -1;
cost[i, j, 1] = -1;
continue;
}
cost[i, j, x % 2] = x;
cost[i, j, 1 - (x % 2)] = -1;
}
}
// Print edge weights
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (cost[i, j, 0] != -1)
Console.Write(cost[i, j, 0] + "\t");
else
Console.Write(cost[i, j, 1] + "\t");
}
Console.WriteLine(" ");
}
Console.ReadLine();
// Find shortest odd and even paths
for (int s = 0; s < 2; s++)
{
for (int k = 0; k < n; k++)
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
for (int u = 0; u <= 1; u++)
for (int v = 0; v <= 1; v++)
{
if (cost[i, k, u] == -1 || cost[k, j, v] == -1)
continue;
int newCost = cost[i, k, u] + cost[k, j, v];
if (newCost < cost[i, j, newCost % 2] || cost[i, j, newCost % 2] == -1)
{
cost[i, j, newCost % 2] = newCost;
// ADDED
path[i, j, newCost % 2] = path[i, k, u].Concat(path[k, j, v]).ToList();
}
}
}
// Print results
Console.WriteLine("\nShortest even paths: ");
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
Console.Write(cost[i, j, 0] + "\t");
Console.WriteLine(" ");
}
Console.WriteLine("\nShortest odd paths:");
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
Console.Write(cost[i, j, 1] + "\t");
Console.WriteLine(" ");
}
Console.WriteLine();
// ADDED
// Example, print shortest odd path between vertices 3 and 1
// This does not print the final q vertex
int p = 3;
int q = 1;
foreach (int index in path[p, q, 1])
Console.Write(index);
Console.ReadLine();
}
}
}
关于c - 使用Floyd-Warshall算法确定 "odd"矩阵,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11900830/