Ultraviolet A 270-lining up

Link:

Http://uva.onlinejudge.org/index.php? Option = com_onlinejudge & Itemid = 8 & category = 113 & page = show_problem & problem = 206

Original question:

''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 cocould only fly over the area once in a straight line, and
She had to fly over as your points as possible. all points were given by means of integer coordinates in a two-dimen1_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 begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. this line is followed by a blank line, and there is also
A blank line between two consecutive inputs.
The input consistsNPairs of integers, where 1 <N<1, 700. each pair of integers is separated by one blank and ended by a new-line character. the list of pairs is ended with an end-of-file character. no pair will occur twice.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
The output consists of one integer representing the largest number of points that all lie on one line.

Sample Input

11 12 23 39 1010 11

Sample output

3

Question:

Give a series of points, and find the maximum number of points that can be connected into a line.

Analysis and Summary:

If two points determine a straight line, you can enumerate all the two points, and then traverse other points in a straight line based on the two points, several points are determined on this line.

To determine whether p1 (x1, Y1), P2 (X2, Y2), and P3 (X3, Y3) are in a straight line, check whether P1, p2 slope and P2, P3 slope is equal, mathematical formula is (x1-x2)/(y1-y2) = (x2-x3)/(y2-y3), but the direct division compared to the slope of the accuracy will be lost, so using the diagonal line to multiply, you can convert this formula into (x1-x2) * (y2-y3) = (y1-y2) * (x2-x3 ).

Then the brute force enumeration is performed.

Code:

/* * UVa:  270 - Lining Up * Time: 0.888s * Author: D_Double * */#include<cstdio>#include<cmath>#include<algorithm>#define MAXN 705using namespace std;struct Node{    int x,y;}arr[MAXN];int nIndex;char str[1000];inline void input(){    nIndex=0;    while(gets(str)){        if(!str[0])break;        sscanf(str,"%d%d",&arr[nIndex].x,&arr[nIndex].y);        ++nIndex;    }}inline bool is_in_line(int X1,int Y1,int X2,int Y2,int X3,int Y3){    return (X1-X2)*(Y3-Y2)-(X3-X2)*(Y1-Y2)==0;}void solve(){    int maxNum=2;    for(int i=0; i<nIndex; ++i){        for(int j=i+1; j<nIndex; ++j){            int cnt=2;            for(int k=j+1; k<nIndex; ++k){                if(is_in_line(arr[i].x,arr[i].y,arr[j].x,arr[j].y,arr[k].x,arr[k].y))                     ++cnt;            }            if(cnt>maxNum) maxNum=cnt;        }    }    printf("%d\n", maxNum);}int main(){    int T;    scanf("%d%*c",&T);    gets(str);    while(T--){        input();        if(nIndex==1)printf("1\n");        else if(nIndex==2)printf("2\n");        else  solve();        if(T) printf("\n");    }    return 0;}

-- The meaning of life is to give it meaning.
 

  Original
  Http://blog.csdn.net/shuangde800
  , By d_double (reprinted, please mark)