二维数组

队友~~高雪彤

一、题目：

输入一个二维整形数组，数组里有正数也有负数。

求所有子数组的和的最大值。

 

二、程序设计思路：

1.先找出二维数组中不包含负数的直接相邻（上下左右相邻）的正整数块。如下图

 

2.然后转化建模，将该“二维表”与“图”转化，转化方法如下：将每个正整数块看做一个“正元素，将每一个负数看做一个“负元素”，将每一个“正元素”和每一个“负元素”分别作为“图”的结点，其中每个正整数块的和的相反数作为相应“正元素”结点的权值，每个负数的值的相反数作为“负元素”结点的权值，具有直接相邻关系的表中元素，具有双向连通性，结点A与结点B的权值的和定义为从结点A到结点B的“代价”，则该题求最大连通子数组的问题成功转化为求图中的最短路径问题，用弗洛伊德算法可以求出图中每对结点之间的最短路径，遍历求出其中的“代价”最小的一组，再还原出其最短路径，即可找出最大连通子图。

 

三、源程序代码

package tt20170227;
import java.util.ArrayList;

public class lala {
    static int sum;
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.print(la333());
    }
 
    public static int la333() {
        int[][] a = { { -8, 9, 2, 6 }, { -5, 8, 6, -9 }, { 6, 1, 6, 7 } };
        int[][] b = new int[3][4];
        // 判断连通性，0为未选中，1为选中，2为连通
        boolean flag = true;
        int sum = 0;

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 4; j++) {
                b[i][j]=0;
                System.out.print(a[i][j] + " ");
                if (a[i][j] >= 0) {
                    b[i][j] = 1;
                }
            }
            System.out.println(" ");
        }

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 4; j++) {
                if (b[i][j] == 1) {
                    switch(i){
                    case 0:
                    {
                        switch(j){
                        case 0:
                        {
                            if (a[i + 1][j] + a[i][j] > 0 && b[i + 1][j] == 0)
                            {
                                b[i + 1][j] = 2;
                            }
                            if (a[i][j + 1] + a[i][j] > 0 && b[i][j + 1] == 0)
                            {
                                b[i][j + 1] = 2;
                            }
                            break;
                        }
                        case 3:
                        {
                            if (a[i + 1][j] + a[i][j] > 0 && b[i + 1][j] == 0) {
                                b[i + 1][j] = 2;
                            }
                            if (a[i][j - 1] + a[i][j] > 0 && b[i][j - 1] == 0) {
                                b[i][j - 1] = 2;
                            }
                            break;
                        }
                        default:
                        {
                            if (a[i + 1][j] + a[i][j] > 0 && b[i + 1][j] == 0) {
                                b[i + 1][j] = 2;
                            }
                            if (a[i][j - 1] + a[i][j] > 0 && b[i][j - 1] == 0) {
                                b[i][j - 1] = 2;
                            }
                            if (a[i][j + 1] + a[i][j] > 0 && b[i][j + 1] == 0) {
                                b[i][j + 1] = 2;
                            }
                            break;
                        }
                        }
                        break;
                    }
                    case 2:
                    {
                        switch(j){
                        case 0:
                        {
                            if (a[i - 1][j] + a[i][j] > 0 && b[i - 1][j] == 0) {
                                b[i - 1][j] = 2;
                            }
                            if (a[i][j + 1] + a[i][j] > 0 && b[i][j + 1] == 0) {
                                b[i][j + 1] = 2;
                            }
                            break;
                        }
                        case 3:
                        {
                            if (a[i - 1][j] + a[i][j] > 0 && b[i - 1][j] == 0) {
                                b[i - 1][j] = 2;
                            }
                            if (a[i][j - 1] + a[i][j] > 0 && b[i][j - 1] == 0) {
                                b[i][j - 1] = 2;
                            }
                            break;
                            }
                        default:
                        {
                            if (a[i - 1][j] + a[i][j] > 0 && b[i - 1][j] == 0) {
                                b[i - 1][j] = 2;
                            }
                            if (a[i][j - 1] + a[i][j] > 0 && b[i][j - 1] == 0) {
                                b[i][j - 1] = 2;
                            }
                            if (a[i][j + 1] + a[i][j] > 0 && b[i][j + 1] == 0) {
                                b[i][j + 1] = 2;
                            }
                            break;
                        }
                        }
                        break;
                    }
                    default:
                    {
                        switch(j){
                        case 0:
                        {
                            if (a[i + 1][j] + a[i][j] > 0 && b[i + 1][j] == 0) {
                                b[i + 1][j] = 2;
                            }
                            if (a[i - 1][j] + a[i][j] > 0 && b[i - 1][j] == 0) {
                                b[i - 1][j] = 2;
                            }
                            if (a[i][j + 1] + a[i][j] > 0 && b[i][j + 1] == 0) {
                                b[i][j + 1] = 2;
                            }
                            break;
                        }
                        case 3:
                        {
                            if (a[i + 1][j] + a[i][j] > 0 && b[i + 1][j] == 0) {
                                b[i + 1][j] = 2;
                            }
                            if (a[i - 1][j] + a[i][j] > 0 && b[i - 1][j] == 0) {
                                b[i - 1][j] = 2;
                            }
                            if (a[i][j - 1] + a[i][j] > 0 && b[i][j - 1] == 0) {
                                b[i][j - 1] = 2;
                            }
                            break;
                            }
                        default:
                        {
                            if (a[i + 1][j] + a[i][j] > 0 && b[i + 1][j] == 0) {
                                b[i + 1][j] = 2;
                            }
                            if (a[i - 1][j] + a[i][j] > 0 && b[i - 1][j] == 0) {
                                b[i - 1][j] = 2;
                            }
                            if (a[i][j - 1] + a[i][j] > 0 && b[i][j - 1] == 0) {
                                b[i][j - 1] = 2;
                            }
                            if (a[i][j + 1] + a[i][j] > 0 && b[i][j + 1] == 0) {
                                b[i][j + 1] = 2;
                            }
                            break;
                        }
                        }
                        break;
                    }
                    }
                }
                System.out.println(a[i][j]);
            }
                
        }
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 4; j++) {
                flag = false;
                if (b[i][j] != 0 && a[i][j] < 0) {
                    b[i][j] = 0;
                    for (int k = 0; k < 3; k++) {
                        for (int q = 0; q < 4; q++) {
                            if (b[k][q] != 0) {
                                if ((b[k + 1][q] <= 0 || b[k + 1][q] > 2) && (b[k - 1][q] <= 0 || b[k - 1][q] > 2)
                                        && (b[k][q + 1] <= 0 || b[k][q + 1] > 2)
                                        && (b[k][q - 1] <= 0 || b[k][q - 1] > 2)) {
                                    flag = true;
                                }
                            }
                        }

                    }
                    if (flag) {
                        b[i][j] = 2;
                    }
                }
            }
        }

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 4; j++) {
                if (b[i][j] != 0) {
                    System.out.print(a[i][j] + " ");
                    sum += a[i][j];
                }
            }
            System.out.println(" ");

        }

        return sum;
    }
}

四、总结：

思路是有的，可是有些还是没有办法具体实现，在课下会多思考，争取实现。