返回二维数组子数组联通和最大

分析

1，开始先将将二维矩阵转换成图的存储形式，当两个相邻的数之间是联通的时，记长度为1，否则就是0；

2，将整个图从每个点都开始遍历一遍，遍历过程中时，当和小于0时断开两点间的路，当和大于最大和时改变最大和的值；

3，取以每个点为起点遍历的和的最大值即时最大联通子数组的和。遍历时，选取已遍历的联通子数组周围最大值遍历。经过重复的几次遍历就可以确定此数组中最大连通数组和了。

题目要求：

放在一个input.txt的文件中

数组里面有正数有负数

返回联通子数组和最大的值

#include<fstream>
#include<iostream>
#define N 100
#include<ctime>
using namespace std;

typedef struct
{
    int d[N];
    int a[N][N];
    int x;
}A;

void set(A &p, int x, int y)                 //x,y分别是行数和列数
{
    p.x = x*y;
    srand((unsigned)time(NULL));
    for (int i = 1; i <= p.x; i++)
    {
        p.d[i] = rand() % 10;
        if (rand() % 2 == 1)
            p.d[i] = p.d[i] * (-1);
    }                                        //随机生成数组的数
    for (int i = 1; i <= p.x; i += y)
    {
        for (int j = i; j <= i + y - 2; j++)
        {
            p.a[j][j + 1] = 1;
            p.a[j + 1][j] = 1;
        }
    }
    for (int i = 1 + y; i<p.x; i += y)
    {
        for (int j = i; j <= i + x - 1; j++)
        {
            p.a[j][j - y] = 1;
            p.a[j - y][j] = 1;
        }
    }                               //将随机生成的一维数组转换成二维的图的形式
}

void bianli(A &p, int v, int visit[], int &b, int &max, int x)
{
    visit[v] = 1;

    max += p.d[v];
    if (max >= b)
        b = max;

    int a = 0, bo = 0;
    for (int w = 1; w <= p.x; w++)
    {
        for (int c = 1; c <= p.x; c++)
        {
            if ((visit[w] == 0) && (p.a[c][w] == 1) && (visit[c] == 1))
            {
                a = w; bo = 1; break;
            }
        }
        if (bo == 1)
            break;
    }
    for (int w = 1; w <= p.x; w++)
    {
        for (int c = 1; c <= p.x; c++)
        {
            if ((visit[w] == 0) && (p.a[c][w] == 1) && (visit[c] == 1))
            {
                if (p.d[a]<p.d[w])
                    a = w;

            }
        }
    }
    if (b + p.d[a]<0)
    {
        p.a[v][a] = 0;
    }
    else
        bianli(p, a, visit, b, max, x);
}
                                                             //遍历

int NoVisit(int visit[], A p)
{
    int k = 0, i;
    for (i = 1; i <= p.x; i++)
    {
        if (visit[i] == 0)
        {
            k = i;
            break;
        }
    }
    return k;
}                                                           //判断图中没有visit的项

int main()
{
    cout << "请输入数组行数和列数：" << endl;
    int x, y;
    cin >> x >> y;
    A p;
    set(p, x, y);
    ofstream fout("D:\\input.txt",ios::binary);
    for (int i = 1; i <= p.x; i++)
    {

        fout  << p.d[i] ;
        if (p.a[i][i + 1] == 1)
            fout << "   ";
        else
            fout << endl;
    }
    int v = 1, b[N] = { 0 }, h = 0;

    for (int i = 1; i <= p.x; i++)
    {
        if (p.d[i]<0)
        {
            b[i] = p.d[i];
        }
        else
        {
            int visit[N] = { 0 };
            int max = 0;
            bianli(p, i, visit, b[i], max, x);
        }
    }

    int max = b[1];
    for (int i = 2; i <= p.x; i++)
    {
        if (b[i]>max)
            max = b[i];
    }
  fout << "最大联通子数组的和为：" << max << endl;
}