• UVA 270 Lining Up (几何 判断共线点)


     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

    题意:给定一些点坐标。求共线点最多的个数。。

    思路:我的做法是暴力枚举。每次先枚举出2个点。作为直线。再去枚举第三个点看在不在直线上。 可以推出一个公式

    y(x1 - x2) = (y1 - y2) * x + y2 * x1 - y1 * x2的时候。为(x,y)在(x1,y1)和(x2,y2)组成的直线上。。不过这样做的话时间复杂度为O(n^3)..中间有个优化。就是如果两个点之前判断连接过了。之后在遇到就直接跳过。。但是依然跑了快2秒。- -


    看网上别人做法有一种时间复杂度为O(n^2logn)的做法。。是每次枚举一个点作为原点。然后把其他点和它的斜率算出来。然后找出这些斜率中相同斜率出现次数最多的作为最大值。感觉不错。

    我的代码:

    #include <stdio.h>
    #include <string.h>
    
    int t;
    int n, i, j, k, l;
    int max, ans;
    int vis[705][705];
    int mark[705];
    char sb[30];
    
    struct Point {
    	int x, y;
    } p[705];
    
    int main() {
    	scanf("%d%*c%*c", &t);
    	while (t --) {
    		n = 0; max = 0;
    		memset(vis, 0, sizeof(vis));
    		while (gets(sb) && sb[0] != '') {
    			sscanf(sb, "%d%d", &p[n].x, &p[n].y);
    			n ++;
    		}
    		for (i = 0; i < n; i ++)
    			for (j = i + 1; j < n; j ++) {
    				if (vis[i][j]) continue;
    				ans = 0;
    				for (k = 0; k < n; k ++) {
    					if (p[k].y * (p[i].x - p[j].x) == (p[i].y - p[j].y) * p[k].x + p[j].y * p[i].x - p[i].y * p[j].x) {
    						mark[ans ++] = k;
    						for (l = 0; l < ans - 1; l ++)
    							vis[k][mark[l]] = vis[mark[l]][k] = 1;
    					}
    				}
    				if (max < ans)
    					max = ans;
    			}
    		printf("%d
    ", max);
    		if (t) printf("
    ");
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/james1207/p/3260319.html
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