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我已经编写了这个程序,该程序使用邻接矩阵实现了具有100个节点的图。我还使用Floyd-Warshall算法为所有100个节点找到所有最短路径对。现在,我想将100 x 100矩阵压缩为10 x 10矩阵,该矩阵仅包含public static final int A = 100
... public static final int W = 66
指定的10个索引中的所有配对最短路径。我应该如何压缩阵列?我已经开始构建一种名为arrayCondenser
的新方法,但有没有更简单的方法来实现这一点?所有对最短路径问题
public class AdjacencyMatrix
{
public static final int NUM_NODES = 100;
public static final int INF = 99999;
public static final int A = 20;
public static final int B = 18;
public static final int C = 47;
public static final int D = 44;
public static final int E = 53;
public static final int F = 67;
public static final int G = 95;
public static final int H = 93;
public static final int I = 88;
public static final int W = 66;
public static boolean even(int num)
{
return num%2==0;
}
public static boolean odd(int num)
{
return num%2==1;
}
public static void initialize(int [][] adjMat, int N)
{
for(int i = 0; i < N; i++)
for(int j = 0; j <N; j++)
adjMat[i][j]=INF;
for(int x = 0; x<N; x++)
{
int row = x/10;
int column = x%10;
if (even(row)) {
if (column!=9)
adjMat[x][x+1]=1;
}
if (odd(row)) {
if (column!=0)
adjMat[x][x-1]=1;
}
if (even(column)){
if (row!=9)
adjMat[x][x+10]=1;
}
if (odd(column)) {
if (row!=0)
adjMat[x][x-10]=1;
}
}
}
public static int[][] floydWarshall(int[][] adjMat, int N)
{
adjMat = new int[N][N];
initialize(adjMat, N);
for(int k = 0; k < N; ++k)
{
for(int i = 0; i < N; ++i)
{
for(int j = 0; j < N; ++j)
{
adjMat[i][j] = Math.min(adjMat[i][j], adjMat[i][k] + adjMat[k][j]);
}
}
}
return adjMat;
}
public static int[][] arrayCondenser(int[][] adjMat, int N)
{
int[] array = {A,B,C,D,E,F,G,H,I,W};
adjMat = floydWarshall(adjMat, N);
return adjMat;
}
public static void printGrid(int[][] adjMat)
{
for (int i=0; i<NUM_NODES; ++i)
{
for (int j=0; j<NUM_NODES; ++j)
{
if (adjMat[i][j]==INF)
System.out.printf("%5s", "INF");
else
System.out.printf("%5d",adjMat[i][j]);
}
System.out.println();
}
}
public static void main(String[] args)
{
int adjMat[][] = new int[NUM_NODES][NUM_NODES];
adjMat = floydWarshall(adjMat, NUM_NODES);
printGrid(adjMat);
System.out.println();
}
}