UVaOJ270 - Lining Up

270 - Lining Up

Time limit: 3.000 seconds

``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 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 consists of N pairs of integers, where 1 < N < 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

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 705;struct Point {    int x, y;}point[MAXN];int nIndex;char str[10000];void init_input() {    nIndex = 0;    while (gets(str)) {        if (! str[0]) break;        sscanf(str, "%d%d", &point[nIndex].x, &point[nIndex].y);        nIndex ++;    }}inline bool is_lined(const int x1, const int y1, const int x2, const int y2, const int x3, const int y3) {    return (x2 - x1) * (y3 - y2) - (x3 - x2) * (y2 - y1) == 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_lined(point[i].x, point[i].y, point[j].x, point[j].y, point[k].x, point[k].y)) {                    cnt ++;                }            }            if (cnt > maxNum) {                maxNum = cnt;            }        }    }    printf("%d\n", maxNum);}int main() {    int T;    scanf("%d%*c", &T);    gets(str);    while (T --) {        init_input();        if (nIndex == 1) {            printf("1\n");        } else if (nIndex == 2){            printf("2\n");        } else {            solve();        }        if (T) {            printf("\n");        }    }    return 0;}