Prime 算法实现

x33g5p2x  于2022-07-11 转载在 其他  
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一 构建后的图

二 实现

package graph.prim;

import java.util.Scanner;

public class Prim {
    static final int INF = 0x3f3f3f3f;
    static final int N = 100;
    // 如果s[i]=true,说明顶点i已加入U
    static boolean s[] = new boolean[N];
    static int c[][] = new int[N][N];
    static int closest[] = new int[N];
    static int lowcost[] = new int[N];

    static void Prim(int n) {
        // 初始时,集合中 U 只有一个元素,即顶点 1
        s[1] = true;
        for (int i = 1; i <= n; i++) {
            if (i != 1) {
                lowcost[i] = c[1][i];
                closest[i] = 1;
                s[i] = false;
            } else
                lowcost[i] = 0;
        }
        for (int i = 1; i < n; i++) {
            int temp = INF;
            int t = 1;
            // 在集合中 V-u 中寻找距离集合U最近的顶点t
            for (int j = 1; j <= n; j++) {
                if (!s[j] && lowcost[j] < temp) {
                    t = j;
                    temp = lowcost[j];
                }
            }
            if (t == 1)
                break; // 找不到 t,跳出循环
            s[t] = true; // 否则,t 加入集合U
            for (int j = 1; j <= n; j++) { // 更新 lowcost 和 closest
                if (!s[j] && c[t][j] < lowcost[j]) {
                    lowcost[j] = c[t][j];
                    closest[j] = t;
                }
            }
        }
    }

    public static void main(String[] args) {
        int n, m, u, v, w;
        Scanner scanner = new Scanner(System.in);
        n = scanner.nextInt();
        m = scanner.nextInt();
        int sumcost = 0;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                c[i][j] = INF;
        for (int i = 1; i <= m; i++) {
            u = scanner.nextInt();
            v = scanner.nextInt();
            w = scanner.nextInt();
            c[u][v] = c[v][u] = w;
        }
        Prim(n);
        System.out.println("数组lowcost:");

        for (int i = 1; i <= n; i++)
            System.out.print(lowcost[i] + " ");

        System.out.println();
        for (int i = 1; i <= n; i++)
            sumcost += lowcost[i];
        System.out.println("最小的花费:" + sumcost);
    }
}

三 测试

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