C#實現的最短路徑分析

複製代碼 代碼如下:using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
class Program
{
static int length = 6;
static string[] shortedPath = new string[length];
static int noPath = 2000;
static int MaxSize = 1000;
static int[,] G =
{
{ noPath, noPath, 10, noPath, 30, 100 },
{ noPath, noPath, 5, noPath, noPath, noPath },
{ noPath, noPath, noPath, 50, noPath, noPath },
{ noPath, noPath, noPath, noPath, noPath, 10 },
{ noPath, noPath, noPath, 20, noPath, 60 },
{ noPath, noPath, noPath, noPath, noPath, noPath }
};
static string[] PathResult = new string[length];

static int[] path1 = new int[length];
static int[,] path2 = new int[length, length];
static int[] distance2 = new int[length];

static void Main(string[] args)
{
int dist1 = getShortedPath(G, 0, 1, path1);
Console.WriteLine("點0到點5路徑:");
for (int i = 0; i < path1.Length; i++)
Console.Write(path1[i].ToString() + " ");
Console.WriteLine("長度:" + dist1);

Console.WriteLine("\r\n-----------------------------------------\r\n");

int[] pathdist = getShortedPath(G, 0, path2);
Console.WriteLine("點0到任意點的路徑:");
for (int j = 0; j < pathdist.Length; j++)
{
Console.WriteLine("點0到" + j + "的路徑:");
for (int i = 0; i < length; i++)
Console.Write(path2[j, i].ToString() + " ");
Console.WriteLine("長度:" + pathdist[j]);
}
Console.ReadKey();

}

//從某一源點出發，找到到某一結點的最短路徑
static int getShortedPath(int[,]G, int start, int end,int [] path)
{
bool[] s = new bool[length]; //表示找到起始結點與當前結點間的最短路徑
int min; //最小距離臨時變數
int curNode=0; //臨時結點，記錄當前正計算結點
int[] dist = new int[length];
int[] prev = new int[length];

//初始結點資訊
for (int v = 0; v < length; v++)
{
s[v] = false;
dist[v] = G[start, v];
if (dist[v] > MaxSize)
prev[v] = 0;
else
prev[v] = start;
}
path[0] = end;
dist[start] = 0;
s[start] = true;
//主迴圈
for (int i = 1; i < length; i++)
{
min = MaxSize;
for (int w = 0; w < length; w++)
{
if (!s[w] && dist[w] < min)
{
curNode = w;
min = dist[w];
}
}

s[curNode] = true;
for (int j = 0; j < length; j++)
if (!s[j] && min + G[curNode, j] < dist[j])
{
dist[j] = min + G[curNode, j];
prev[j] = curNode;
}

}
//輸出路徑結點
int e = end, step = 0;
while (e != start)
{
step++;
path[step] = prev[e];
e = prev[e];
}
for (int i = step; i > step/2; i--)
{
int temp = path[step - i];
path[step - i] = path[i];
path[i] = temp;
}
return dist[end];
}

//從某一源點出發，找到到所有結點的最短路徑
static int[] getShortedPath(int[,] G, int start, int[,] path)
{
int[] PathID = new int[length];//路徑（用編號表示）
bool[] s = new bool[length]; //表示找到起始結點與當前結點間的最短路徑
int min; //最小距離臨時變數
int curNode = 0; //臨時結點，記錄當前正計算結點
int[] dist = new int[length];
int[] prev = new int[length];
//初始結點資訊
for (int v = 0; v < length; v++)
{
s[v] = false;
dist[v] = G[start, v];
if (dist[v] > MaxSize)
prev[v] = 0;
else
prev[v] = start;
path[v,0] = v;
}

dist[start] = 0;
s[start] = true;
//主迴圈
for (int i = 1; i < length; i++)
{
min = MaxSize;
for (int w = 0; w < length; w++)
{
if (!s[w] && dist[w] < min)
{
curNode = w;
min = dist[w];
}
}

s[curNode] = true;

for (int j = 0; j < length; j++)
if (!s[j] && min + G[curNode, j] < dist[j])
{
dist[j] = min + G[curNode, j];
prev[j] = curNode;
}

}
//輸出路徑結點
for (int k = 0; k < length; k++)
{
int e = k, step = 0;
while (e != start)
{
step++;
path[k, step] = prev[e];
e = prev[e];
}
for (int i = step; i > step / 2; i--)
{
int temp = path[k, step - i];
path[k, step - i] = path[k, i];
path[k, i] = temp;
}
}
return dist;

}

}
}