10245 - The Closest Pair Problem

描述：刚开始直接超时，后来改了一点，居然过了
#include <cstdio>
#include <cmath>
#define N 10010
double num
[2];
int main()
{
// freopen("a.txt","r",stdin);
int n;
while(scanf("%d",&n)!=EOF)
{
if(!n) break;
double sum=100000000,count;
for(int i=0; i<n; i++)
{
scanf("%lf %lf",&num[i][0],&num[i][1]);
for(int j=0;j<i;j++)
{
double a=num[i][0]-num[j][0],b=num[i][1]-num[j][1];
count=a*a+b*b;
if(count<sum) sum=count;
}
}
if(sum>=100000000) printf("INFINITY\n");
else printf("%.4lf\n",sqrt(sum));
}
return 0;
}