C 使用Floyd Warshall算法确定;“奇数”;矩阵
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。我试图做的不是简单地寻找最短的路径,我希望最短的路径也是一个均匀的权重 例如,这是Floyd Warshall算法的一个简单实现:C 使用Floyd Warshall算法确定;“奇数”;矩阵,c,algorithm,path-finding,floyd-warshall,C,Algorithm,Path Finding,Floyd Warshall,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。我试图做的不是简单地寻找最短的路径,我希望最短的路径也是一个均匀的权重 例如,这是Floyd Warshall算法的一个简单实现: #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
#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);
}
}
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”)
我的代码目前所做的是,它获得了在每个单独的“奇数”和“偶数”矩阵中表示的最佳权重
我缺乏理解的是,当最优值位于相反的矩阵(奇数/偶数)中时,“奇数”和“偶数”矩阵如何得出它们的非最优值。在我看来,必须有某种循环或递归才能做到这一点,但我不知道如何做到这一点。在C中不是,但这不应该是一个问题。我认为F-W需要两种修改才能获得最短奇偶路径:
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();
}
}
}
使用系统;
使用System.Collections.Generic;
使用System.Linq;
使用系统文本;
命名空间控制台应用程序5
{
班级计划
{
静态void Main(字符串[]参数)
{
int n=5;
//生成图形
随机r=新随机(1);
//增加
列表[,]路径=新列表[n,n,2];
整数[,]成本=新整数[n,n,2];
对于(int i=0;i 对于(int u=0;u,我不确定这是否有效,但值得一试:将权重拆分为distOdd
和dist偶数
矩阵,然后运行三个嵌套循环。在每个步骤上执行四个任务:(1)检查[I][j]
和[j][k]
处的两个偶数路径是否可以改善[I][k]
处的偶数路径,(2)查看两条奇数路径是否可以改善[i][k]
处的偶数路径,以及(3)查看[i][j]
处的奇数路径和[j][k]
处的偶数路径是否可以改善[i][k]
处的奇数路径,以及(4)查看[i][j]
处的偶数路径和[j][k]
处的奇数路径是否可以改善[i][k]处的奇数路径
@dasblinkenlight我的问题是算法已经找到了最优化的路径。例如,在处理第3列期间,2->1
中的最佳路径的权重为“2”,但由于2是一个偶数,它不知怎的找到了数字“5”,该数字似乎是通过节点3循环一次得到的,并且由于算法m不支持递归性,我要么需要以某种方式添加它,要么找到另一种方法来找到它。我有点困惑三维是如何花费[,]
正在运行。难道不需要两个不同的矩阵来获得奇数和偶数路径吗?它们实际上只是两个二维矩阵,因为第三维是2。cost[x,y,0]是偶数矩阵,cost[x,y,1]是奇数矩阵。我用它来代替两个变量,这样我就可以编写cost[I,j,newCost%2]之类的东西=newCost而不是“if偶数更新偶数矩阵,else更新奇数矩阵”此方法是否允许您使用后续方法来追溯您所走的路径?据我所知(我还无法测试),它应该能够准确地找到我所要求的,但如果我想
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
-------------------------------
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();
}
}
}