poj2606

题意：给定平面内的一些点，求平面内最多有几个点共线。

分析：枚举两个点确定一条线段，把所有线段按照斜率排序，然后整理斜率相等的即可获得答案。

View Code

#include < iostream >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cmath >
#include < algorithm >
using namespace std;

#define eps 1.0e-8
#define maxn 205

struct XPoint
{
int x, y;
} point[maxn];

struct Line
{
int a, b;
double k;
bool end;
} line[maxn * maxn];

int n, ncount;

bool operator < ( const Line & a, const Line & b)
{
if (abs(a.k - b.k) > eps && ! a.end && ! b.end)
return a.k < b.k;
if (a.end != b.end)
return a.end < b.end;
if (a.a != b.a)
return a.a < b.a;
return a.b < b.b;
}

double getk(XPoint & a, XPoint & b)
{
return (b.y - a.y) * 1.0 / (b.x - a.x);
}

int main()
{
// freopen("t.txt", "r", stdin);
scanf( " %d " , & n);
for ( int i = 0 ; i < n; i ++ )
scanf( " %d%d " , & point[i].x, & point[i].y);
for ( int i = 0 ; i < n - 1 ; i ++ )
for ( int j = i + 1 ; j < n; j ++ )
{
line[ncount].a = i;
line[ncount].b = j;
if (point[i].x == point[j].x)
line[ncount].end = true ;
else
{
line[ncount].end = false ;
line[ncount].k = getk(point[i], point[j]);
}
ncount ++ ;
}
sort(line, line + ncount);
int start = 0 ;
int ans = 0 ;
for ( int i = 1 ; i < ncount; i ++ )
{
if ( ! ((line[i].end && line[start].end) || (( ! line[i].end && ! line[start].end) && (line[i].k - line[start].k)
< eps)) || line[i].a != line[start].a)
{
start = i;
continue ;
}
if (i - start > ans)
{
ans = i - start;
}
}
printf( " %d\n " , ans + 2 );
return 0 ;
}