Saddle Point (5 points)

Title Contents:

Given an n * n matrix A. A saddle point is a matrix position (i, j), the element at that position is the maximum number of the i-th row, the minimum number of j-th column. A matrix may not have a saddle point.

Your task is to find the saddle point A.

Input formats:

The first input line is a positive integer n, (1 <= n <= 100), then there are n rows, each row has n integers, with one or more spaces between two integers on the same row.

Output formats:

Matrix input, if found saddle point, it outputs the index. Subscript two numbers, the first number is the line number and the second number is the row number, from 0 and starts counting.

If not, then output

NO

Title given to ensure that the data multiple saddle point does not appear.

Sample input:

4 

1 7 4 1 

4 8 3 6 

1 6 1 2 

0 7 8 9

Sample output:

2 1

#include<stdio.h>
#define nn 100
int main()
{
	int n;
	scanf("%d",&n);
	int max[nn] = {0};//储存每行最大的数
	int min[nn] = {0};//储存每列最小的数
	int a[nn][nn] = {0};//二维数组
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			scanf("%d",&a[i][j]);
		}
	} 
	for(int i=0;i<n;i++){//找到每行最大的数，每列最小的数
		int big = 1,small = 100;
		for(int j=0;j<n;j++){
				if(a[j][i]<small){
					small = a[j][i];
				}
				if(a[i][j]>big){
					big = a[i][j];
				}
		}
		max[i] = big;
		min[i] = small;
	}
	int flag = 1;//标志
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			if(a[i][j]==max[i]&&a[i][j]==min[j]){//做判断是否为鞍点
				flag = 0;
				printf("%d %d",i,j);
			}
		}
	} 
	if(flag){//如果没有，则输出NO
		printf("NO");
	}
}