UVa 270:lining up

Link:

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

Original title:

' How am I I ever going to solve this problem? ' said the pilot.

Indeed, the pilot is not facing a 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 Pos Sible. All points were given by means of the integer coordinates in a two-dimensional spaces. The pilot wanted to know the largest number of points from the given set, all lie on one line. Can you write a program so calculates this number?

Your program has to be efficient!

Input

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The input begins with a single positive integer in a line by itself indicating the number of cases following, each of them as described below. This are followed by a blank line, and there are also a blank line between two consecutive. The input consists of N pairs of integers, where 1 < N < 700. Each pair of integers was 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 would occur twice.

Output

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

Sample Input

1

1 1
2 2 3 3 9
ten
10 11

Sample Output

3

The main effect of the topic:

Give a series of points, ask the maximum number of points can be connected to a line.

Analysis and Summary:

Two points to determine a line, then you can enumerate all two points of the situation, and then in accordance with the two points determined by a straight line, and then traverse the other points, to determine a few points in this line.

The way to determine whether a three point P1 (x1,y1), p2 (x2,y2), P3 (X3,Y3) is a straight line is to see whether the slope of the P1,P2 is equal to the slope of the P2,P3, the mathematical formula is (X1-X2)/(Y1-y2) = (x2-x3)/(Y2-Y3), But the direct division compares the slope the accuracy to have the loss, therefore uses the diagonal to multiply the rule, may convert this equation to (x1-x2) * (y2-y3) = (y1-y2) * (X2-X3).

Then there is the violent enumeration.

Code:

* * uva:270-lining up * time:0.888s * author:d_double * * * * * * * * */#include <cstdio> #include <cmath&gt  
;  
#include <algorithm> #define MAXN 705 using 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; }

Author: csdn Blog shuangde800