uva1606 Amphiphilic Carbon Molecules

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题目的主要思想在于依次选定一个节点，然后再选定剩下的另外一个节点，然后判断以这两个节点确定的直线作为边界，相应的黑色节点的个数以及白色节点的个数的最大值，这里也使用了一个很重要的方法，将黑色节点的位置进行中心对称，这样就将求解白色节点以及黑色节点的个数统一起来了，直接求出节点数然后取最大值就行了，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;const int maxn = 1005;int N;typedef struct node{double x, y;double rad;}node;bool compare(const node a,const node b){return a.rad < b.rad;}int x[maxn], y[maxn], color[maxn], ans;bool isLeft(node a,node b){//由a旋转到breturn a.x*b.y - a.y*b.x >= 0;}void deal(){for (int i = 0; i < N; i++){vector<node> v;for (int j = 0; j < N; j++){if (j != i){node temp;temp.x = x[j] - x[i];temp.y = y[j] - y[i];if (color[j]){temp.x = -temp.x, temp.y = -temp.y;}temp.rad = atan2(temp.y, temp.x);v.push_back(temp);}}sort(v.begin(), v.end(), compare);int L = 0, R = 0;int amount = 2, up = v.size();while (L < up){if (L == R){R = (R + 1) % up;amount++;}while (R != L&&isLeft(v[L], v[R])){R = (R + 1) % up;amount++;}L++;amount--;ans = max(ans, amount);}}}int main(){ while (cin >> N){if (N == 0) break;for (int i = 0; i < N; i++){cin >> x[i] >> y[i] >> color[i];}ans = -1;deal();cout << ans << endl;}return 0;}

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