弗洛伊德算法
- 求最短路径算法( 各个节点到其他节点的最短路径)
- 时间复杂度为n^3
- 通过三个for循环:
一、介绍
二、思想
三、代码实现
package com.achang.algorithm;
import java.util.Arrays;
public class Floyd {
public static void main(String[] args) {
char[] vertex = {'a', 'b', 'c', 'd', 'e', 'f', 'g'};
final int N = 65535;//不可达常量数字
int[][] matrix = {
// a b c d e f g
{0, 5, 7, N, N, N, 2},//a
{5, 0, N, 9, N, N, 3},//b
{7, N, 0, N, 8, N, N},//c
{N, 9, N, 0, N, 4, N},//d
{N, N, 8, N, 0, 5, 4},//e
{N, N, N, 4, 5, 0, 6},//f
{2, 3, N, N, 4, 6, 0}//g
};
Graph1 graph1 = new Graph1(vertex.length, matrix, vertex);
graph1.showPreArr();
graph1.showDisArr();
System.out.println("==========");
graph1.floyd();
graph1.showPreArr();
graph1.showDisArr();
}
}
//图结构
class Graph1 {
private char[] vertex;//存放节点的数组
private int[][] dis;//保存从各个节点出发到其他节点的距离,最后的结果在这里
private int[][] pre;//保存到达目标节点的前驱节点
public Graph1(int length, int[][] matrix, char[] vertex) {
this.vertex = vertex;
this.dis = matrix;
this.pre = new int[length][length];
//对pre数组进行初始化,存放的是前驱节点的下标
for (int i = 0; i < length; i++) {
Arrays.fill(pre[i], i);
}
}
public void showPreArr() {
System.out.println("遍历pre数组===");
for (int[] pr : pre) {
System.out.println(Arrays.toString(pr));
}
}
public void showDisArr() {
System.out.println("遍历dis数组===");
for (int[] di : dis) {
System.out.println(Arrays.toString(di));
}
}
public void show() {
for (int i = 0; i < pre.length; i++) {
//输出dis数组
for (int i1 = 0; i1 < dis.length; i1++) {
System.out.print(dis[i][i1] + "");
}
System.out.println();
//输出pre数组
for (int i1 = 0; i1 < dis.length; i1++) {
System.out.print(pre[i][i1] + "");
}
System.out.println();
}
}
//弗洛伊德算法,求各个节点到其他节点的最短路径
public void floyd(){
int len = 0;//变量保存距离
//对中间节点遍历,k就是中间节点的下标
for (int k = 0; k < dis.length; k++) {
//对下标为j节点的开始节点遍历[a,b,c,d,e,f,g]
for (int i = 0; i < dis.length; i++) {
//对终点节点遍历,k就是终点节点的下标
for (int j = 0; j < dis.length; j++) {
len = dis[i][k]+dis[k][j];//求出从i节点出发,经过为k的中间节点,到达最终节点j节点的距离
//dis[i][j]直连距离:下标i为的出发节点,到下标j的终点节点的距离
if (len < dis[i][j]){
dis[i][j] = len;
pre[i][j] = pre[k][j];
}
}
}
}
}
}