标签:print algorithm private system 数组 alt i++ 开始 fill
看一个应用场景和问题:
战争时期,胜利乡有7个村庄(A, B, C, D, E, F, G) ,现在有六个邮差,从G点出发,需要分别把邮件分别送到 A, B, C , D, E, F 六个村庄
各个村庄的距离用边线表示(权) ,比如 A – B 距离 5公里
问:如何计算出G村庄到 其它各个村庄的最短距离?
如果从其它点出发到各个点的最短距离又是多少?
/**
* 存放顶点的数组
*/
private String[] data;
/**
* 保存,从各个顶点出发到其它顶点的距离
*/
private int[][] dis;
/**
* 保存到达目标顶点的前驱顶点
*/
private int[][] pre;
public FloydAlgorithm(String[] data, int[][] dis) {
this.data = data;
this.dis = dis;
this.pre = new int[data.length][data.length];
for (int i = 0; i < pre.length; i++) {
Arrays.fill(pre[i], i);
}
}
public void floydAlgorithm() {
int len = 0;
// 1. 对中间顶点遍历, k就是中间顶点的下标
for (int k = 0; k < dis.length; k++) {
// 2. 从i顶点开始出发
for (int i = 0; i < dis.length; i++) {
// 3. 到达j顶点
for (int j = 0; j < dis.length; j++) {
len = dis[i][k] + dis[k][j];
if (len < dis[i][j]) {
dis[i][j] = len;
pre[i][j] = k;
}
}
}
}
}
public void show(){
for (int i = 0; i < dis.length; i++) {
for (int j = 0; j < dis[i].length; j++) {
System.out.print(data[i]+"==>"+data[j]+"最短路径为:"+dis[i][j]+"\t");
}
System.out.println();
}
}
public static void main(String[] args) {
// 测试看看图是否创建成功
String[] data = {"A", "B", "C", "D", "E", "F", "G"};
//创建邻接矩阵
int[][] matrix = new int[data.length][data.length];
final int N = 65535;
matrix[0] = new int[]{0, 5, 7, N, N, N, 2};
matrix[1] = new int[]{5, 0, N, 9, N, N, 3};
matrix[2] = new int[]{7, N, 0, N, 8, N, N};
matrix[3] = new int[]{N, 9, N, 0, N, 4, N};
matrix[4] = new int[]{N, N, 8, N, 0, 5, 4};
matrix[5] = new int[]{N, N, N, 4, 5, 0, 6};
matrix[6] = new int[]{2, 3, N, N, 4, 6, 0};
FloydAlgorithm floydAlgorithm = new FloydAlgorithm(data, matrix);
floydAlgorithm.floydAlgorithm();
floydAlgorithm.show();
}
标签:print algorithm private system 数组 alt i++ 开始 fill
原文地址:https://www.cnblogs.com/MND1024/p/11611593.html