Amphiphilic Carbon Molecules UVA

题目传送门

题意：平面上有n个点，每一个点为白点或者黑点。现在要放置一个隔板，使得隔板的一侧白点加上另一侧黑点的总数最大。

思路：这个题借鉴了这位博主的做法http://blog.csdn.net/u014800748/article/details/47952253

#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <fstream>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <sstream>#include <stack>#include <string>#include <vector>#define MAXN 1010#define MAXE 210#define INF 10000000#define MOD 1000000007#define LL long long#define pi acos(-1.0)using namespace std;struct Point {  int x;  int y;  double rad;  bool operator<(const Point &p) const { return rad < p.rad; }} p[MAXN], point[MAXN];int color[MAXN];int n;bool check(Point p1, Point p2) { return p1.x * p2.y - p1.y * p2.x >= 0; }int solve() {  if (n <= 2)    return 2;  int ans = 0;  for (int i = 0; i < n; ++i) {    int num = 0;    for (int j = 0; j < n; ++j) {      if (j != i) {        p[num].x = point[j].x - point[i].x;        p[num].y = point[j].y - point[i].y;        if (color[j]) {          p[num].x = -p[num].x;          p[num].y = -p[num].y;        }        p[num].rad = atan2(p[num].y, p[num].x);        num++;      }    }    sort(p, p + num);    int L = 0, R = 0, cnt = 2;    while (L < num) {      if (L == R) {        R = (R + 1) % num;        cnt++;      }      while (L != R && check(p[L], p[R])) {        R = (R + 1) % num;        cnt++;      }      L++;      cnt--;      ans = max(ans, cnt);    }  }  return ans;}int main() {  std::ios::sync_with_stdio(false);  while (cin >> n && n) {    for (int i = 0; i < n; ++i)      cin >> point[i].x >> point[i].y >> color[i];    cout << solve() << endl;  }  return 0;}/*30 0 00 1 02 2 140 0 00 4 04 0 01 2 17-1 0 01 2 12 3 02 1 10 3 11 4 0-1 2 00*/