uva 270 Lining Up
``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
1 1 1 2 2 3 3 9 10 10 11
Sample Output
3
题目大意:给出一系列的点, 求最多有多少个点能连成一条线。
解题思路:当三点共线时:A(X1,Y1),B(X2,Y2),C(X3,Y3) ————> (Y2-Y1)/(X2-X1) = (Y3-Y2)/(X3-X2)。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include<algorithm>
using namespace std;
int X[1005], Y[1005], cnt, Max;
int CNT(int a, int b, int c) { //计算在这条线上的点的数量
int cnt2 = 0;
for (int i = c + 1; i < cnt; i++) {
if (a * (Y[i] - Y[c]) == b * (X[i] - X[c])) cnt2++;
}
return cnt2;
}
int findMax() {
for (int i = 0; i < cnt; i++) {
for (int j = i + 1; j < cnt; j++) {
int temp = CNT(X[j] - X[i], Y[j] - Y[i], j) + 2; //本身两个点
Max = max(Max, temp);
}
}
}
char s[1005];
int main() {
int T;
scanf("%d%*c%*c", &T); //注意读空行
while (T--) {
cnt = 0;
while (gets(s)) {
if (!s[0] && cnt) break;
sscanf(s, "%d%d", &X[cnt], &Y[cnt]);
cnt++;
}
Max = -1;
findMax();
printf("%d\n", Max);
if (T) printf("\n");
}
return 0;
}