返回一个最大联通子数组的和

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

typedef struct
{
int dian
;
int xian

;
int dianx, xianx;
}A;

void set(A &shu, int x, int y)
{
shu.dianx = x*y;
srand((unsigned)time(NULL));
for (int i = 1; i <= shu.dianx; i++)
{
shu.dian[i] = rand() % 10;
if (rand() % 2 == 1)
shu.dian[i] = shu.dian[i] * (-1);
}
for (int i = 1; i <= shu.dianx; i += y)
{
for (int j = i; j <= i + y - 2; j++)
{
shu.xian[j][j + 1] = 1;
shu.xian[j + 1][j] = 1;
}
}
for (int i = 1 + y; i<shu.dianx; i += y)
{
for (int j = i; j <= i + x - 1; j++)
{
shu.xian[j][j - y] = 1;
shu.xian[j - y][j] = 1;
}
}
}
void output(A shu)
{
for (int i = 1; i <= shu.dianx; i++)
{
cout << shu.dian[i] ;
if (shu.xian[i][i + 1] == 1)
cout << "  ";
else
cout << endl;
}
}
void bianli(A &shu, int v, int visit[], int &b, int &max, int x)
{
visit[v] = 1;

max += shu.dian[v];
if (max >= b)
b = max;

int a = 0, bo = 0;
for (int w = 1; w <= shu.dianx; w++)
{
for (int c = 1; c <= shu.dianx; c++)
{
if ((visit[w] == 0) && (shu.xian[c][w] == 1) && (visit[c] == 1))
{
a = w; bo = 1; break;
}
}
if (bo == 1)
break;
}
for (int w = 1; w <= shu.dianx; w++)
{
for (int c = 1; c <= shu.dianx; c++)
{
if ((visit[w] == 0) && (shu.xian[c][w] == 1) && (visit[c] == 1))
{
if (shu.dian[a]<shu.dian[w])
a = w;
}
}
}
if (b + shu.dian[a]<0)
{
shu.xian[v][a] = 0;
}
else
bianli(shu, a, visit, b, max, x);
}

int NoVisit(int visit[], A shu)
{
int k = 0, i;
for (i = 1; i <= shu.dianx; i++)
{
if (visit[i] == 0)
{
k = i;
break;
}
}
return k;
}

int main()
{
cout << "请输入数组行列数：" << endl;
int x, y;
cin >> x >> y;
A shu;
set(shu, x, y);
output(shu);

int v = 1, b
= { 0 }, h = 0;
for (int i = 1; i <= shu.dianx; i++)
{
if (shu.dian[i]<0)
{
b[i] = shu.dian[i];
}
else
{
int visit
= { 0 };
int max = 0;
bianli(shu, i, visit, b[i], max, x);
}
}

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

题目要求：

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。



设计思想：这道题目感觉很难，第一次设想是求出每一行最大子数组的和，同时求出他们的坐标，比较他们是否在同一行，后来发现很难实现

通过图的遍历可以查找

结果：并不是自己所做，题目还是偏难。