C++找出二维数组的所有鞍点。所谓鞍点就是在该位置上的元素在该行上最大，在该列上最小。

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define M 5
#define N 4
struct COORDINATE{
int x;
int y;
int sts;
}std[M];
int max_ary(int ary[M][N],int row,int length)
{
int i;
int tmp;
int rst;
rst = 0;
tmp = ary[row][0];
for(i=1;i<length;i++){
if(tmp<ary[row][i]){
rst = i;
tmp = ary[row][i];
}
}
return rst;
}
int min_ary(int ary[M][N],int cel,int length)
{
int i;
int tmp;
int rst;
rst = 0;
tmp = ary[0][cel];
for(i=1;i<length;i++){
if(tmp>ary[i][cel]){
rst = i;
tmp = ary[i][cel];
}
}
return rst;
}
int main(void) {
int i;
int x,y;
int ary[M][N]={{2,2,3,4},
{2,3,4,5},
{6,4,0,5},
{2,5,6,7},
{2,6,7,8}};

memset(std,0,sizeof(std));

for(i=0;i<M;i++){
std[i].x = max_ary(ary,i,N);
std[i].y = i;
}
for(i=0;i<M;i++){
if(std[i].y==min_ary(ary,std[i].x,M)){
std[i].sts = 1;
}
}
for(i=0;i<M;i++){
if(std[i].sts == 1){
x = std[i].x;
y = std[i].y;
printf("the Row=%d,Cel=%d,data=%d\n",y,x,ary[y][x]);
}
}
return EXIT_SUCCESS;
}

Answer 2

随手写了个，可能有误，思路供你参考，希望有所收获。

void count(char [M][N]org) //M行，N列
{
int row[N],col[M];
int rCount = cCount = 0;
for(int i = 0 ; i < M-1 ; i++ ) //找出每行最大的
{
int cNum = 0;
int temp = [i][0];
for(int j = 1 ; j < N-1 ; j++ )
{
if(org[i][j]>temp)
{
temp = org[i][j];
cNum = j;
}
}
row[rCount++] = cNum;
}

for(int j = 0 ; j < N-1 ; j++ ) //找出列最小的
{
int rNum = 0;
int temp = [0][j];
for(int i = 1 ; i < M-1 ; i++ )
{
if(org[i][j]<temp)
{
temp = org[i][j];
rNum = j;
}
}
col[cCount++] = rNum;
}

for(int i = 0 ; i < M ; i++) //找出行中最大的恰好是列中最小的
if(col[row[i]] == i)
cout<<"第"+i+"行,第"+row[i]+"列为鞍点";

}