Hdu 1432 Lining Up【简单几何】

Lining Up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1343 Accepted Submission(s): 385

Problem 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

The input consists of multiple test cases, and each case begins with a single positive integer on a line by itself indicating the number of points, followed by N pairs of integers, where 1 < N < 700. Each pair of integers is separated by one blank and ended by a new-line character. No pair will occur twice in one test case.

Output

For each test case, the output consists of one integer representing the largest number of points that all lie on one line, one line per case.

Sample Input

51 12 23 39 1010 11

Sample Output

3

题意：

给出一些点的坐标，问最多有多少点在同一条直线上......

题解：

每次固定一个点，枚举其他各个点和这个点连成的直线，直线斜率相等的肯定在同一条直线上

比赛做的时候，考虑的有点多，把整个直线的方程都求出来了，后来想想，只需要求斜率就好了.....

/*http://blog.csdn.net/liuke19950717*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=705;struct point {int y,x;}x[maxn];struct line{int kase;//标记是否有斜率 double k;}y[maxn*maxn];int cmp(line a,line b){if(a.kase==b.kase){return a.k<b.k;}return a.kase<b.kase;}line cal(point a,point b){line tp={0,0};int ty=a.y-b.y,tx=a.x-b.x;if(tx==0)//斜率不存在 {tp.kase=0;}else//斜率存在，求直线 {tp.kase=1;tp.k=ty*1.0/tx;}return tp;}bool ok(line a,line b){return a.kase==b.kase&&a.k==b.k;}int slove(int n){int ans=0;for(int i=0;i<n;++i){int cnt=0;for(int j=0;j<n;++j){if(i==j){continue;}y[cnt]=cal(x[i],x[j]);++cnt;}sort(y,y+cnt,cmp);int tp=1;for(int j=1;j<cnt;++j){if(ok(y[j],y[j-1])){++tp;}else{tp=1;}ans=max(ans,tp);}}return ans+1;}int main(){int n;while(~scanf("%d",&n)){for(int i=0;i<n;++i){scanf("%d%d",&x[i].y,&x[i].x);}printf("%d\n",slove(n));}return  0;}