Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
思路：
这道题的数据好像很弱，我只用了最普通的方法居然可以AC。
对于每个点，求其他点到他连线的斜率，如果斜率相同，就表示在同一条直线上，用hash记录斜率和点的数目的关系。复杂度是O(n^2)的。
最开始出错是因为没有考虑重复点的情况，加上这方面判断就AC了。
代码：
1 int maxPoints(vector<Point> &points) { 2 int n = points.size(); 3 if(n <= 2) 4 return n; 5 map<float, int> slope; 6 map<pair<int, int>, bool> visited; 7 int res = 1; 8 int duplicate; 9 for(int i = 0; i < n; i++){ 10 slope.clear(); 11 if(visited.find(pair<int, int>(points[i].x, points[i].y)) != visited.end()) 12 continue; 13 visited[pair<int, int>(points[i].x, points[i].y)] = true; 14 duplicate = 0; 15 int tres = 1; 16 for(int j = 0; j < n; j++){ 17 if(j == i) 18 continue; 19 if(points[i].x == points[j].x && points[i].y == points[j].y){ 20 duplicate++; 21 continue; 22 } 23 float tmp; 24 if(points[i].x == points[j].x) 25 tmp = INT_MAX; 26 else 27 tmp = (float)(points[i].y-points[j].y)/(float)(points[i].x-points[j].x); 28 if(slope.find(tmp) == slope.end()) 29 slope[tmp] = 2; 30 else 31 slope[tmp]++; 32 tres = tres > slope[tmp] ? tres : slope[tmp]; 33 } 34 res = res > tres+duplicate ? res : tres+duplicate; 35 } 36 return res; 37 }