#include <stdio.h>
#include <math.h>
void main()
{
    int a[2][3];
    int i,j,h=0,l=0,max1,max2,r,h1=0,l1=0;
    for(i=0;i<2;i++)
        for(j=0;j<3;j++)
            scanf("%d",&a[i][j]);
    max1=a[i][j];
    for(i=0,j=0;j<3;j++)
            if(max1>a[i][j])
            {
                max1=a[i][j];
                h=i;
                l=j;
            }
            printf(" h=%d l=%d\n",h,l);
            max2=a[i][j];
            for(i=1,j=0;j<3;j++)
                if(max2>a[i][j])
                {
                    max2=a[i][j];
                    h1=i;
                    l1=j;
                }
                printf(" h1=%d l1=%d\n",h1,l1);
                printf(" max1=%d max2=%d\n",max1,max2);
                for(i=0,j=l;i<2;i++)
                    if(max1<a[i][j])
                        max1=0;
                    for(i=0,j=l1;i<2;i++)
                        if(max2<a[i][j])
                            max2=0;
                        printf(" max1=%d max2=%d\n",max1,max2);
                        if(max1==0)
                            r=max2;
                        else if(max2==0)
                            r=max1;
                        else if(max1==0&&max2==0)
                            r=0;
                        printf("r=%d\n",r);
}


看了下答案的算法，只求所有列最大的就可以了吗？比如1 2 3
                                                  4 5 6  鞍点不是4吗

在矩阵中，一个数在所在行中是最大值，在所在列中是最小值，则被称为鞍点。所以不是了哟！