【求帮忙改代码】这是一个完整的代码，要把它改成一个类
这是Flody算法……，我想把它改成一个类，一个只输入两个地点就出结果的……例：输入1，6这两个点，显示的是
from 1 to 6 的路径是15
经过的点是1，2，3，6；
public class FLOYD {
int[][] length = null;// 任意两点之间路径长度
int[][][] path = null;// 任意两点之间的路径
public FLOYD(int[][] G) {
int MAX = 100;
int row = G.length;// 图G的行数
int[][] spot = new int[row][row];// 定义任意两点之间经过的点
int[] onePath = new int[row];// 记录一条路径
length = new int[row][row];
path = new int[row][row][];
for (int i = 0; i < row; i++)
// 处理图两点之间的路径
for (int j = 0; j < row; j++) {
if (G[i][j] == 0)
G[i][j] = MAX;// 没有路径的两个点之间的路径为默认最大
if (i == j)
G[i][j] = 0;// 本身的路径长度为0
}
for (int i = 0; i < row; i++)
// 初始化为任意两点之间没有路径
for (int j = 0; j < row; j++)
spot[i][j] = -1;
for (int i = 0; i < row; i++)
// 假设任意两点之间的没有路径
onePath[i] = -1;
for (int v = 0; v < row; ++v)
for (int w = 0; w < row; ++w)
length[v][w] = G[v][w];
for (int u = 0; u < row; ++u)
for (int v = 0; v < row; ++v)
for (int w = 0; w < row; ++w)
if (length[v][w] > length[v][u] + length[u][w]) {
length[v][w] = length[v][u] + length[u][w];// 如果存在更短路径则取更短路径
spot[v][w] = u;// 把经过的点加入
}
for (int i = 0; i < row; i++) {
// 求出所有的路径
int[] point = new int[1];
for (int j = 0; j < row; j++) {
point[0] = 0;
onePath[point[0]++] = i;
outputPath(spot, i, j, onePath, point);
path[i][j] = new int[point[0]];
for (int s = 0; s < point[0]; s++)
path[i][j][s] = onePath[s];
}
}
}
void outputPath(int[][] spot, int i, int j, int[] onePath, int[] point) {// 输出i
// 到j
// 的路径的实际代码,point[]记录一条路径的长度
if (i == j)
return;
if (spot[i][j] == -1)
onePath[point[0]++] = j;
// System.out.print(" "+j+" ");
else {
outputPath(spot, i, spot[i][j], onePath, point);
outputPath(spot, spot[i][j], j, onePath, point);
}
}
public static void main(String[] args) {
int data[][] = {
{ 0, 27, 44, 17, 11, 27, 42, 0, 0, 0, 20, 25, 21, 21, 18, 27, 0 },// x1
{ 27, 0, 31, 27, 49, 0, 0, 0, 0, 0, 0, 0, 52, 21, 41, 0, 0 },// 1
{ 44, 31, 0, 19, 0, 27, 32, 0, 0, 0, 47, 0, 0, 0, 32, 0, 0 },// 2
{ 17, 27, 19, 0, 14, 0, 0, 0, 0, 0, 30, 0, 0, 0, 31, 0, 0 },// 3
{ 11, 49, 0, 14, 0, 13, 20, 0, 0, 28, 15, 0, 0, 0, 15, 25, 30 },// 4
{ 27, 0, 27, 0, 13, 0, 9, 21, 0, 26, 26, 0, 0, 0, 28, 29, 0 },// 5
{ 42, 0, 32, 0, 20, 9, 0, 13, 0, 32, 0, 0, 0, 0, 0, 33, 0 },// 6
{ 0, 0, 0, 0, 0, 21, 13, 0, 19, 0, 0, 0, 0, 0, 0, 0, 0 },// 7
{ 0, 0, 0, 0, 0, 0, 0, 19, 0, 11, 20, 0, 0, 0, 0, 33, 21 },// 8
{ 0, 0, 0, 0, 28, 26, 32, 0, 11, 0, 10, 20, 0, 0, 29, 14, 13 },// 9
{ 20, 0, 47, 30, 15, 26, 0, 0, 20, 10, 0, 18, 0, 0, 14, 9, 20 },// 10
{ 25, 0, 0, 0, 0, 0, 0, 0, 0, 20, 18, 0, 23, 0, 0, 14, 0 },// 11
{ 21, 52, 0, 0, 0, 0, 0, 0, 0, 0, 0, 23, 0, 27, 22, 0, 0 },// 12
{ 21, 21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 27, 0, 0, 0, 0 },// 13
{ 18, 41, 32, 31, 15, 28, 0, 0, 0, 29, 14, 0, 22, 0, 0, 11, 0 },// 14
{ 27, 0, 0, 0, 25, 29, 33, 0, 33, 14, 9, 14, 0, 0, 11, 0, 9 },// 15
{ 0, 0, 0, 0, 30, 0, 0, 0, 21, 13, 20, 0, 0, 0, 0, 9, 0 } // 16
};
for (int i = 0; i < data.length; i++)
for (int j = i; j < data.length; j++)
if (data[i][j] != data[j][i])
return;
FLOYD test=new FLOYD(data);
for (int i = 0; i < data.length; i++)
for (int j = i; j < data[i].length; j++) {
System.out.println();
System.out.print("From " + i + " to " + j + " path is: ");
for (int k = 0; k < test.path[i][j].length; k++)
System.out.print(test.path[i][j][k] + " ");
System.out.println();
System.out.println("From " + i + " to " + j + " length :"
+ test.length[i][j]);
}
}
}
--------------------编程问答-------------------- 这个已经封装的挺好的了，了不起再增加个函数：
public int path(int posFrom, int posTo) {
System.out.print("From " + posFrom + " to " + posTo + "\tstep:" + this.length[posFrom][posTo] + "\tPATH: ");
for (int k = 0; k < this.path[posFrom][posTo].length; k++)
System.out.print(this.path[posFrom][posTo][k] + " ");
System.out.println();
return this.length[posFrom][posTo];
}
调用时就两句话：
FLOYD test=new FLOYD(data); // 创建该对象，data是你要提供的图；
test.path(0, 5); // 从0到5
补充：Java , Java相关
