从城市A到B,C,D,E城市，每个城市都要去一次，求最短通路，用JAVA怎么做啊，我弄了一个代码，但是只能求A到每个城市的距离，求不了通路
public class Dijkstra {
static int M=10000;
public static void main(String[] args) {
int[][] weight1 = {//邻接矩阵
{0,3,2000,7,M},
{3,0,4,2,M},
{M,4,0,5,4},
{7,2,5,0,6},
{M,M,4,6,0}
};
int[][] weight2 = {
{0,130,244,126,141},
{130,0,274,M,186},
{244,274,0,M,M},
{126,M,M,0,58},
{141,186,M,58,0}
};
int start=0;
int[] shortPath = Dijsktra(weight2,start);
for(int i = 0;i < shortPath.length;i++)
System.out.println("从"+start+"出发到"+i+"的最短距离为："+shortPath[i]);
}
public static int[] Dijsktra(int[][] weight,int start){
int n = weight.length;
int[] shortPath = new int[n];
String[] path=new String[n];
for(int i=0;i<n;i++)
path[i]=new String(start+"-->"+i);
int[] visited = new int[n];
shortPath[start] = 0;
visited[start] = 1;
for(int count = 1;count <= n - 1;count++)
{
int k = -1;
int dmin = Integer.MAX_VALUE;
for(int i = 0;i < n;i++)
{
if(visited[i] == 0 && weight[start][i] < dmin)
{
dmin = weight[start][i];
k = i;
}
}
System.out.println("k="+k);
shortPath[k] = dmin;
visited[k] = 1;
for(int i = 0;i < n;i++)
{
if(visited[i] == 0 && weight[start][k] + weight[k][i] < weight[start][i]){
weight[start][i] = weight[start][k] + weight[k][i];
path[i]=path[k]+"-->"+i;
}
}
}
for(int i=0;i<n;i++)
System.out.println("从"+start+"出发到"+i+"的最短路径为："+path[i]);
System.out.println("=====================================");
return shortPath;
}
}
遍历一下，你把你这个程序写成函数，然后一个城市到其他城市的距离都可以算出来，然后求一下最小值即可