【1122】简单鞍点

描述：

矩阵中比上下两个数都大且比左右两个数都小数称为“鞍点”。求输入的矩阵中鞍点的个数。
In matrix the number who is bigger then the upper and lower two numbers, andsmaller then the left and right two numbers, called "saddle point".count input matrix saddle-point number.

输入：

输入的第一行是两个整数m、n（2<n,m<100），代表矩阵有m行n列；
接下来的m行每行有n个正整数。
input two integers m and n (2<n,m<100), which represents a matrix of mrows n columns;
Next m lines each line have n positive integer.

输出：

输出鞍点的个数。格式是printf("%d\n", count);
Output the number of the saddle-point

输入样例：

3 4
1 2 13 4
6 5 8 7
4 3 12 1

#include<stdio.h>int main(){int m,n,count=0;scanf("%d%d",&m,&n);int f[m][n],i,j;for(i=0;i<m;i++){for(j=0;j<n;j++){scanf("%d",&f[i][j]);}}for(i=1;i<m-1;i++){for(j=1;j<n-1;j++){if(f[i][j]>f[i-1][j]&&f[i][j]>f[i+1][j]&&f[i][j]<f[i][j-1]&&f[i][j]<f[i][j+1])count=count+1;}}printf("%d\n", count);return 0;}