hdoj--1432--Lining Up（暴力模拟）

Lining Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1274    Accepted Submission(s): 366

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

 

Source

ACM暑期集训队练习赛（四）

 

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题意：n个点，然后求同一条线上数量最多的点的个数，，，，就是那样吧，求点的个数

思路：先对点排序，求出所有直线的k还有b，然后寻找最多的相同的k，b个数，然后去重

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#include<iostream>#define INF 0x3f3f3f3f#define ull unsigned long long#define ll long long#define IN __int64#define N 710#define M 100000000using namespace std;struct zz{    int x;    int y;}p[N];int cmp(zz a,zz b){    if(a.x==b.x)        return a.y<b.y;    return a.x<b.x;}struct ss{    double k;    double b;}pp[N*N];bool cmpp(ss a,ss b){    if(a.k==b.k)        return a.b<b.b;    return a.k<b.k;}int main(){    int n,i,j,k;    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<n;i++)            scanf("%d%d",&p[i].x,&p[i].y);        sort(p,p+n,cmp);        int kk=0;        int km=0;        for(i=0;i<n-1;i++)        {            for(j=i+1;j<n;j++)            {                int x=p[j].x-p[i].x;                int y=p[j].y-p[i].y;                pp[kk].k=y*1.0/x;                pp[kk].b=p[i].y-pp[kk].k*p[i].x;                kk++;            }        }        sort(pp,pp+kk,cmpp);        int m=1,mm=1;        ss f=pp[0];        for(i=1;i<kk;i++)        {            if(fabs(pp[i].k-f.k)<1e-6&&fabs(f.b-pp[i].b)<1e-6)            {                m++;                mm=max(mm,m);            }            else            {                f=pp[i];                m=1;            }        }        if(n==1)            printf("1\n");        else if(n==2)            printf("2\n");        else        {            if(mm==3)                printf("%d\n",mm);            else            {                mm*=2;                for(int i=1;i<=n;i++)                 {                    if(i*(i+1)==mm)                    {                        printf("%d\n",i+1);                        break;                    }                }            }        }    }    return 0;}