java算法篇---銀行家算法

銀行家算法是一種最有代表性的避免死鎖的算法。在避免死鎖方法中允許進程動態地申請資源，但系統在進行資源分配之前，應先計算此次分配資源的安全性，若分配不會導致系統進入不安全狀態，則分配，否則等待。

package com.zc.manythread;
import javax.swing.JOptionPane;
/*一共有５個進程需要請求資源，有３類資源*/
public class Bank {
    // 每個進程所需要的最大資源數
    public static int MAX[][] = { { 7, 5, 3 }, { 3, 2, 2 }, { 9, 0, 2 },
            { 2, 2, 2 }, { 4, 3, 3 } };
    // 系統擁有的初始資源數
    public static int AVAILABLE[] = { 10, 5, 7 };
    // 系統已給每個進程分配的資源數
    public static int ALLOCATION[][] = { { 0, 0, 0 }, { 0, 0, 0 }, { 0, 0, 0 },
            { 0, 0, 0 }, { 0, 0, 0 } };
    // 每個進程還需要的資源數
    public static int NEED[][] = { { 7, 5, 3 }, { 3, 2, 2 }, { 9, 0, 2 },
            { 2, 2, 2 }, { 4, 3, 3 } };
    // 每次申請的資源數
    public static int Request[] = { 0, 0, 0 };
    // 進程數與資源數
    public static int M = 5, N = 3;
    int FALSE = 0;
    int TRUE = 1;
    public void showdata() {
        int i, j;
        System.out.print("系統可用的資源數爲:/n");
        for (j = 0; j < N; j++) {
            System.out.print("資源" + j + ":" + AVAILABLE[j] + " ");
        }
        System.out.println();
        System.out.println("各進程還需要的資源量:");
        for (i = 0; i < M; i++) {
            System.out.print("進程" + i + ":");
            for (j = 0; j < N; j++) {
                System.out.print("資源" + j + ":" + NEED[i][j] + " ");
            }
            System.out.print("/n");
        }
        System.out.print("各進程已經得到的資源量: /n");
        for (i = 0; i < M; i++) {
            System.out.print("進程");
            System.out.print(i);
            for (j = 0; j < N; j++) {
                System.out.print("資源" + j + ":" + ALLOCATION[i][j] + " ");
            }
            System.out.print("/n");
        }
    }
    // 分配資源，並重新更新各種狀態
    public void changdata(int k) {
        int j;
        for (j = 0; j < N; j++) {
            AVAILABLE[j] = AVAILABLE[j] - Request[j];
            ALLOCATION[k][j] = ALLOCATION[k][j] + Request[j];
            NEED[k][j] = NEED[k][j] - Request[j];
        }
    };
    // 回收資源，並重新更新各種狀態
    public void rstordata(int k) {
        int j;
        for (j = 0; j < N; j++) {
            AVAILABLE[j] = AVAILABLE[j] + Request[j];
            ALLOCATION[k][j] = ALLOCATION[k][j] - Request[j];
            NEED[k][j] = NEED[k][j] + Request[j];
        }
    };
    // 釋放資源
    public void free(int k) {
        for (int j = 0; j < N; j++) {
            AVAILABLE[j] = AVAILABLE[j] + ALLOCATION[k][j];
            System.out.print("釋放" + k + "號進程的" + j + "資源!/n");
        }
    }
    public int check0(int k) {
        int j, n = 0;
        for (j = 0; j < N; j++) {
            if (NEED[k][j] == 0)
                n++;
        }
        if (n == 3)
            return 1;
        else
            return 0;
    }
    // 檢查安全性函數
    public int chkerr(int s) {
        int WORK;
        int FINISH[] = new int[M], temp[] = new int[M];// 保存臨時的安全進程序列
        int i, j, k = 0;
        for (i = 0; i < M; i++)
            FINISH[i] = FALSE;
        for (j = 0; j < N; j++) {
            WORK = AVAILABLE[j]; // 第j個資源可用數
            i = s;
            // 判斷第i個進程是否滿足條件
            while (i < M) {
                if (FINISH[i] == FALSE && NEED[i][j] <= WORK) {
                    WORK = WORK + ALLOCATION[i][j];
                    FINISH[i] = TRUE;
                    temp[k] = i;
                    k++;
                    i = 0;
                } else {
                    i++;
                }
            }
            for (i = 0; i < M; i++)
                if (FINISH[i] == FALSE) {
                    System.out.print("/n系統不安全!!! 本次資源申請不成功!/n");
                    return 1;
                }
        }
        System.out.print("/n經安全性檢查，系統安全，本次分配成功。/n");
        System.out.print("本次安全序列：");
        for (i = 0; i < M - 1; i++) {
            System.out.print("進程" + temp[i] + "->");
        }
        System.out.print("進程" + temp[M - 1]);
        System.out.println("/n");
        return 0;
    }
    /**
     * @param args
     */
    public static void main(String[] args) {
        int i = 0, j = 0;
        int flag = 1;
        Bank bank = new Bank();
        bank.showdata();
        while (flag == 1) {
            i = -1;
            while (i < 0 || i >= M) {
                String str = JOptionPane.showInputDialog("請輸入需申請資源的進程號（從0到"
                        + (M - 1) + "，否則重輸入!）");
                i = Integer.parseInt(str);
                if (i < 0 || i >= M)
                    System.out.println("輸入的進程號不存在，重新輸入!/n");
            }
            System.out.print("請輸入進程" + i + "申請的資源數/n");
            for (j = 0; j < N; j++) {
                String str = JOptionPane.showInputDialog("資源" + j + ":");
                Request[j] = Integer.parseInt(str);
                if (Request[j] > NEED[i][j]) {
                    System.out.print("進程" + i + "申請的資源數大於進程" + i + "還需要" + j
                            + "類資源的資源量!申請不合理，出錯!請重新選擇!/n");
                    flag = 0;
                    break;
                } else {
                    if (Request[j] > AVAILABLE[j]) {
                        System.out.print("進程" + i + "申請的資源數大於系統可用" + j
                                + "類資源的資源量!申請不合理，出錯!請重新選擇!/n");
                        flag = 0;
                        break;
                    }
                }
            }
            if (flag == 1) {
                bank.changdata(i);
                int chkerr = bank.chkerr(i);
                if (chkerr == 1) {
                    bank.rstordata(i);
                    bank.showdata();
                } else {
                    bank.showdata();
                    int check = bank.check0(i);
                    if (check == 1) {
                        System.out.print("進程" + i + "已經完成，系統將其所佔用資源釋放/n");
                        bank.free(i);
                    }                    
                }
            } else {
                bank.showdata();
            }
            System.out.println("/n");
            String str = JOptionPane
                    .showInputDialog("是否繼續銀行家算法演示,按'1'鍵繼續,按'0'鍵退出演示");
            flag = Integer.parseInt(str);
        }
    }
}