返回一個最大聯通子數組的和

標籤：

#include<iostream>#include<ctime>using namespace std;#define N 100typedef struct{    int dian[N];    int xian[N][N];    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 << "請輸入數組行列數：" << endl;    int x, y;    cin >> x >> y;    A shu;    set(shu, x, y);    output(shu);    int v = 1, b[N] = { 0 }, h = 0;    for (int i = 1; i <= shu.dianx; i++)    {        if (shu.dian[i]<0)        {            b[i] = shu.dian[i];        }        else        {            int visit[N] = { 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 << "最大聯通子數組的和為：" << max << endl;}

題目要求：

輸入一個二維整形數組，數組裡有正數也有負數。

求所有子數組的和的最大值。

 

設計思想：這道題目感覺很難，第一次設想是求出每一行最大子數組的和，同時求出他們的座標，比較他們是否在同一行，後來發現很難實現

通過圖的遍曆可以尋找

結果：並不是自己所做，題目還是偏難。

返回一個最大聯通子數組的和