UVa 270 - Lining Up

来源：互联网发布：钓鱼岛知乎编辑：程序博客网时间：2024/06/04 17:49

题目：给你平面上n个点，求最多有几个点共线。

分析：计算几何。枚举基准点，其他点按基准点极角排序（斜率排序），然后枚举相邻直线判断统计即可。

说明：时间复杂度O（n*n*lg（n））。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;typedef struct pnode{double x,y;}point;point  P[701];double S[701];int main(){int  t;char buf[100];scanf("%d",&t);getchar();gets(buf);while (t --) {int count = 0;while (gets(buf) && buf[0]) {sscanf(buf, "%lf%lf", &P[count].x,&P[count].y);count ++;}int max = 0;for (int i = 0 ; i < count ; ++ i) {int save = 0;for (int j = 0 ; j < count ; ++ j) {if (j == i) continue;if (P[i].x == P[j].x)S[save ++] = 1e1000;elseS[save ++] = (P[j].y-P[i].y)/(P[j].x-P[i].x);}sort(S, S+save);int now = 1;for (int j = 1 ; j < save ; ++ j)if (S[j] == S[j-1])now ++;else {if (max < now) max = now;now = 1;}if (max < now) max = now;}printf("%d\n",max+1);if (t) printf("\n");}    return 0;}

0 0