- 首先转为图,从一个顶点出发,找到相邻的最短路径的顶点,然后再找前俩个节点的所有路径中最短路径的顶点,依次类推
public class PrimAlgorithm {
public static void main(String[] args) {
char[] data = {'A', 'B', 'C', 'D', 'E', 'F', 'G'};
int verxs = data.length;
int[][] weight = {
{10000, 5, 7, 10000, 10000, 10000, 2},
{5, 10000, 10000, 9, 10000, 10000, 3},
{7, 10000, 10000, 10000, 8, 10000, 10000},
{10000, 9, 10000, 10000, 10000, 4, 10000},
{10000, 10000, 8, 10000, 10000, 5, 4},
{10000, 10000, 10000, 4, 5, 10000, 6},
{2, 3, 10000, 10000, 4, 6, 10000}
};
MinTree mt = new MinTree();
MGraph graph = mt.createGraph(verxs, data, weight);
mt.show(graph);
mt.prim(graph, 0);
}
}
class MinTree {
public MGraph createGraph(int verxs, char[] data, int[][] weight) {
MGraph graph = new MGraph(verxs, data, weight);
return graph;
}
public void show(MGraph graph) {
for (int i = 0; i < graph.weight.length; i++) {
System.out.println(Arrays.toString(graph.weight[i]));
}
}
public void prim(MGraph graph, int v) {
int[] visited = new int[graph.verxs];
visited[v] = 1;
int minWeight = 1000;
int minX = 0, minY = 0;
for (int i = 1; i < graph.verxs; i++) {
for (int j = 0; j < graph.verxs; j++) {
for (int k = 0; k < graph.verxs; k++) {
if (visited[j] == 1 && visited[k] == 0 && graph.weight[j][k] < minWeight) {
minWeight = graph.weight[j][k];
minX = j;
minY = k;
}
}
}
System.out.println("点:" + graph.data[minX] + "====点:" + graph.data[minY] + "---weight:"
+ graph.weight[minX][minY]);
visited[minY] = 1;
minWeight = 10000;
}
}
}
class MGraph {
int verxs;
char[] data;
int[][] weight;
public MGraph(int verxs, char[] data, int[][] weight) {
this.verxs = verxs;
this.data = data;
this.weight = weight;
}
}
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