【算法设计与分析基础】8、穷举 旅行商问题

package cn.xf.algorithm.ch03;

import java.util.LinkedList;

import org.apache.commons.lang.StringUtils;
import org.junit.Test;

/**
*
* 功能：穷举 旅行商问题
* @author xiaofeng
* @date 2017年4月26日
* @fileName TravelingSalesmanProblem.java
*
*/
public class TravelingSalesmanProblem {

/**
* 递归穷举所有路径，路径可以但是和不对！！！,求高手指点？？？？
* @param data
* @param points
* @param paths
* @param count
* @param pre
* @param current
*/
public static void getPathAndCount(int data[][], LinkedList<Character> points, String paths, String pathCount,
int count, char pre, char current) {
if (points.size() == 0) {
System.out.println(
paths.toString() + " count:" + pathCount.substring(0, pathCount.length() - 2) + "=" + count);
} else {
int len = points.size();
String seq = "abcd";
// 保存当前路径节点
// char oldpre = new Character(current);
pre = new Character(current);
pathCount = new String(pathCount);
for (int i = 0; i < len; ++i) {
// 这个是浅拷贝
LinkedList<Character> newPoints = (LinkedList<Character>) points.clone();
// pre = current;
current = new Character(newPoints.get(i));
newPoints.remove(i);

int index1 = seq.indexOf(pre);
int index2 = seq.indexOf(current);

String newPaths = new String(paths);
if (pre != current) {
newPaths += "->" + current;
if (index1 != -1 && index2 != -1) {
count += data[index1][index2];
pathCount += data[index1][index2] + " + ";
}
}
getPathAndCount(data, newPoints, newPaths, pathCount, count, pre, current);
}
}
}

/**
* 换个说法，就是吧这个几个城市的全排列枚举出来
* @param data
* @param points
* @param paths
* @param pathCount
* @param count
* @param pre
* @param current
*/
public static void getPathAndCount2(int data[][], LinkedList<Character> points, String paths) {
if(points.size() == 0){
String seq = "abcd";
String resultPath[] = paths.split("->");
//在这个时候进行统计数据和
String countPath = "";
int count = 0;
String pre = resultPath[0];
String cur;
for(int j = 1; j < resultPath.length; ++j){
if(!StringUtils.isEmpty(countPath)){
//如果是空，那么说明没有加入数据
countPath += " + " ;
}
//统计和路径
cur = resultPath[j];
countPath += data[seq.indexOf(pre)][seq.indexOf(cur)];
//求和
count += data[seq.indexOf(pre)][seq.indexOf(cur)];
pre = resultPath[j];
}

System.out.println(paths + " count:" + countPath + " = " + count);
return;
}

int len = points.size();
for(int i = 0; i < len; ++i){
//递归到节点全部加入,这里不能直接把原本的节点加入，应该是对一个新的链表操作，深拷贝
LinkedList<Character> newPoints = (LinkedList<Character>) points.clone();
String newPaths = new String(paths);
Character cur = points.get(i);
if(StringUtils.isNotEmpty(paths)){
//如果路径非空，说明已经有节点
newPaths += "->" + cur;
} else {
newPaths += cur;
}
newPoints.remove(i);	//移除当前加入路径的节点
getPathAndCount2(data, newPoints, newPaths);
}
}

@Test
public void testGetPathAndCount(){
int data[][] = {{0,2,5,7},{2,0,8,3},{5,8,0,1},{7,3,1,0}};
LinkedList<Character> points = new LinkedList<Character>();
points.add('a');points.add('b');points.add('c');points.add('d');
String paths = "";
String pathCount = "   ";
int count = 0; char pre = '0'; char current = '0';
TravelingSalesmanProblem.getPathAndCount2(data, points, paths);
//		TravelingSalesmanProblem.getPathAndCount(data, points, paths, pathCount, count, pre, current);

}

public static void main(String[] args) {
String seq = "abcd";
int data[][] = {{0,2,5,7},{2,0,8,3},{5,8,0,1},{7,3,1,0}};
System.out.println(seq.indexOf('a'));
}
}