找出二维数组的“鞍点”

 1//找出一个二维数组的“鞍点”，即该位置上的元素在该行上最大，在该列上最小。也可能没有鞍点。
 2 
 3//(1)二维数组有鞍点：
 4 int a[3][4]={{9,80,205,40},{90,-60,96,1},{210,-3,101,89}};
 5 
 6 //(2)二维数组没有鞍点：
 7 int b[4][4]={{9,80,205,40},{90,-60,196,1},{210,-3,101,89},{45,54,156,7}};
 8 
 9 AnDian(a, 3);
10 AnDian(b, 4);
11 
12/*
13查找鞍点函数
14注意：二维数组作参数，一定要给出列的维度
15*/
16void AnDian(int arr[][4], int len)
17{
18 int i, j, k;
19 int rowMaxIndex, iRAn=-1, iCAn=-1, iAn=-1;
20 int iTemp;
21 bool bAn = true;
22 
23 printf("\n---------------------------------\n\n");
24 for (i=0; i<len; i++)
25 {
26  for (j=0; j<4; j++)
27  {
28   printf("%-5d  ",arr[i][j]);
29  }
30  printf("\n\n");
31 }
32 //------------------------------
33 //find max number in row
34 for (i=0; i<len; i++)//Row
35 {
36  bAn = true;
37  iTemp = arr[i][0];
38  rowMaxIndex = 0;
39  for (j=0; j<4; j++)//Col
40  {
41   if (iTemp < arr[i][j])
42   { 
43    iTemp = arr[i][j];
44    rowMaxIndex = j;
45   }
46  }
47 
48  //judge wether iTemp is min number in Cols
49  for (k=0; k<len; k++)//Row
50  {   
51   if (iTemp > arr[k][rowMaxIndex])
52   {    
53    bAn = false;
54    break;
55   }   
56  } 
57  if (bAn)
58  {
59   iRAn = i;
60   iCAn = rowMaxIndex;
61   iAn = iTemp;
62  }
63 }// for (i=0; i<len; i++)//Row
64 if (iRAn != -1)
65 {
66  printf("鞍点位置：(%d, %d) 鞍点值：%d\n\n", iRAn, iCAn, iAn);
67 }
68 else
69 {
70  printf("没有鞍点\n\n");
71 }
72}