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设计思路:
首先将二维数组中正整数的值找出来,之后找到每个正整数上下左右加起来为正的负数。之后判断是否联通,将小的负数排除掉,最后留下的是二维整数数组中最大联通子数组。
程序代码:
1 #include <iostream> 2 #include <time.h> 3 #define M 3 4 #define N 5 5 using namespace std; 6 7 void main() 8 { 9 int a[M][N] = {0},b[M][N]={0};//判断联通性,0为未选中,1为选中,2为连通 10 bool flg = 0; //判断是否有1存在,存在为O。 11 int sum = 0; //最后和 12 13 srand(unsigned((int)time(0))); 14 for (int i = 0;i < M;i++) 15 { 16 for (int j = 0;j < N;j++) 17 { 18 a[i][j] = rand()%50 - 20; 19 cout << a[i][j] << "\t"; 20 if (a[i][j] >= 0) 21 { 22 b[i][j] = 1; 23 } 24 } 25 cout << endl; 26 } 27 cout << endl; 28 29 for (int i = 0;i < M;i++) 30 { 31 for (int j = 0;j < N;j++) 32 { 33 if (b[i][j] == 1) 34 { 35 if (a[i+1][j] + a[i][j] > 0 && b[i+1][j] == 0) 36 { 37 b[i+1][j] = 2; 38 } 39 if (a[i-1][j] + a[i][j] > 0 && b[i-1][j] == 0) 40 { 41 b[i-1][j] = 2; 42 } 43 if (a[i][j-1] + a[i][j] > 0 && b[i][j-1] == 0) 44 { 45 b[i][j-1] = 2; 46 } 47 if (a[i][j+1] + a[i][j] > 0 && b[i][j+1] == 0) 48 { 49 b[i][j+1] = 2; 50 } 51 } 52 } 53 } 54 55 for (int i = 0;i < M;i++) 56 { 57 for (int j = 0;j < N;j++) 58 { 59 flg = 0; 60 if (b[i][j] != 0 && a[i][j] < 0) 61 { 62 b[i][j] = 0; 63 for (int k = 0;k < M;k++) 64 { 65 for (int l = 0;l < N;l++) 66 { 67 if (b[k][l] != 0) 68 { 69 if ((b[k+1][l] <= 0 || b[k+1][l] > 2)&& 70 (b[k-1][l] <= 0 || b[k-1][l] > 2)&& 71 (b[k][l+1] <= 0 || b[k][l+1] > 2)&& 72 (b[k][l-1] <= 0 || b[k][l-1] > 2)) 73 { 74 flg = 1; 75 } 76 } 77 } 78 } 79 if (flg) 80 { 81 b[i][j] = 2; 82 } 83 } 84 } 85 } 86 87 for (int i = 0;i < M;i++) 88 { 89 for (int j = 0;j < N;j++) 90 { 91 if (b[i][j] != 0) 92 { 93 cout << a[i][j] << "\t"; 94 sum += a[i][j]; 95 } 96 else 97 { 98 cout << "**" << "\t"; 99 } 100 } 101 cout << endl; 102 } 103 104 cout << "sum = " << sum << endl; 105 }
结果截图:
总结:
这个开始想思路的时候就感觉很难,编程的时候发现的确如此。实现的结果并不理想。虽然结果大部分会对,可是还是有一部分bug,本来是在往后补漏洞,修改,可是越补越多,感觉不应该这样。目前只能这样了。
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原文地址:http://www.cnblogs.com/zrdm/p/4464113.html