编程基本功——寻找矩阵中的“鞍点”

一、问题分析

    在所在行中最大，在所在列中最小。所以一个矩阵最多只有一个鞍点。

     所以可以逐行寻找鞍点。先找出某行中最大的元素，再将该元素与同列中其他元素进行比较，若该元素同时是所在列中最小的元素，则它就是鞍点。

二、源码

   1: #include <stdio.h>

   2:  

   3: int GetSaddlePoint(int *x, int *y, int (*A)[5], int m, int n)

   4: {

   5:     int max, i, j, k, flag;

   6:     for (i = 0; i < m; ++i)

   7:     {

   8:         max = 0;

   9:         flag = 1;

  10:         for (j = 1; j < n; ++j)

  11:         {

  12:             if (A[i][j] > A[i][max])

  13:             {

  14:                 max = j;

  15:             }

  16:         }

  17:  

  18:         for (k = 0; k < n; ++k)

  19:         {

  20:             if (A[i][max] == A[i][k] && max != k)

  21:             {

  22:                 flag = 0;

  23:                 break;

  24:             }

  25:         }

  26:  

  27:         if (flag == 1)

  28:         {

  29:             for (k = 0; k < m; ++k)

  30:             {

  31:                 if (A[k][max] <= A[i][max] && k != i)

  32:                 {

  33:                     flag = 0;

  34:                     break;

  35:                 }

  36:             }

  37:         }

  38:  

  39:         if (flag == 1)

  40:         {

  41:             *x = i;

  42:             *y = max;

  43:             return 1;

  44:         }

  45:     }

  46:  

  47:     return 0;

  48: }

  49:  

  50: int main()

  51: {

  52:     int A[5][5], i, j, x, y;

  53:     printf("please input some digit into the 5*5 matrix\n");

  54:     for (i = 0; i < 5; ++i)

  55:     {

  56:         for (j = 0; j < 5; ++j)

  57:         {

  58:             scanf("%d", &A[i][j]);

  59:         }

  60:     }

  61:  

  62:     if (GetSaddlePoint(&x, &y, A, 5, 5))

  63:     {

  64:         printf("\n SaddlePoint is (%d , %d)\n", x+1, y+1);

  65:     }

  66:     else

  67:     {

  68:         printf("there is no SaddlePoint\n");

  69:     }

  70:     return 0;

  71: }