用产生式得出最短巡查路线的距离的算法是什么？-灵析社区
拽嘻嘻比如5个地点ABCDE，两两之间可直达，彼此距离也知道，从A出发巡查其他4个地点后回到A，问最短的距离是多少？ 我做的是选择题，只有答案，不知道步骤如何来的。 
花影流年package com.wm.course.utils;
import java.util.ArrayList;
import java.util.List;
public class TSP {
// 距离矩阵
private static final int[][] distances = {
{0, 7, 10, 5, 10},
{7, 0, 10, 7, 10},
{10, 10, 0, 6, 8},
{5, 7, 6, 0, 9},
{10, 10, 8, 9, 0}
};
// 地点数量
private static final int N = distances.length;
// 保存最短路径
private static List shortestPath = new ArrayList();
// 保存最短路径的距离
private static int shortestDistance = Integer.MAX_VALUE;
public static void main(String[] args) {
List path = new ArrayList();
path.add(0); // 从A开始
boolean[] visited = new boolean[N];
visited[0] = true;
tsp(path, visited, 0, 0);
System.out.println("最短路径: " + shortestPath);
System.out.println("最短距离: " + shortestDistance);
}
private static void tsp(List path, boolean[] visited, int currentCity, int currentDistance) {
if (path.size() == N) {
// 回到起点A
currentDistance += distances[currentCity][0];
if (currentDistance (path);
shortestPath.add(0); // 添加起点A
}
return;
}
for (int city = 0; city < N; city++) {
if (!visited[city]) {
visited[city] = true;
path.add(city);
tsp(path, visited, city, currentDistance + distances[currentCity][city]);
path.remove(path.size() - 1);
visited[city] = false;
}
}
}
}