Lining Up
Time Limit: 2000MS
Memory Limit: 32768K
Total Submissions: 25990
Accepted: 8140
Description
"How am I ever going to solve this problem?" said the pilot.
Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number?
Your program has to be efficient!
Input
Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.
Output
output one integer for each input case ,representing the largest number of points that all lie on one line.
Sample Input
5
1 1
2 2
3 3
9 10
10 11
0
Sample Output
3
Source
East Central North America 1994甚至不用搜索..O(n^3)也能过
//Writer:GhostCai && His Yellow Duck
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
struct point{
int x,y;
}a[705];
inline int read_d(){
int i=0;
char c;
while(c=getchar(),c<'0'||c>'9');
while(c>='0'&&c<='9'){
i=i*10+c-'0';
c=getchar();
}
return i;
}
inline bool cmp(point x,point y){
return x.x < y.x ;
}
int main(){
int n;
while(1){
scanf("%d",&n);
if(n==0) return 0;
for(int i=1;i<=n;i++){
scanf("%d%d",&a[i].x ,&a[i].y );
}
int ans=1;
sort(a+1,a+1+n,cmp);
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++) {
int sum=2;
for(int k=j+1;k<=n;k++){
if((a[i].x-a[j].x)*(a[j].y-a[k].y)==(a[j].x-a[k].x)*(a[i].y-a[j].y)) sum++;
}
ans=max(ans,sum);
}
}
printf("%d
",ans);
}
return 0;
}