标签：style blog color 使用 io for div line
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
思路：对于两个点(x1,y1)和(x2,y2)，其所确定的直线方程为：(x1-x2)*y + (y2-y1)*x + x2*y1-x1*y2 = 0，即：a*y + b*x + c = 0。
使用类似广度优先遍历的思想，Q[i]保存了第i+1个点与前i个点中的每一个点所确定的直线、以及前i+1个点中经过该直线的点的数目，Q[i][0]表示第i+1个点已经出现的次数。
依次处理输入序列的每一个元素，对于当前元素i，依次访问每一个Q[k] (k = i-1, i-2, ..., 0)，遍历Q[k]找到经过点i和点k所确定的直线的最大的点的数目，保存在Q[i]中。
从后往前访问Q[k]的过程中，在第一次遇见点i和点k重合时，更新Q[i][0]的次数值。
1 class Line 2 { 3 public: 4 Line(): a(0), b(0), c(0), cnt(1) {} 5 ~Line(){} 6 public: 7 int a, b, c; 8 int cnt; 9 }; 10 11 class Solution { 12 public: 13 int maxPoints( vector<Point> &points ) { 14 if( points.size() <= 2 ) { return points.size(); } 15 vector<Line>* Q = new vector<Line>[points.size()]; 16 Q[0].resize(1); 17 int max_num = 2; 18 Line optimal; 19 for( int i = 1; i < points.size(); ++i ) { 20 Q[i].resize(1); 21 for( int j = i-1; j >= 0; --j ) { 22 optimal.a = points[i].x-points[j].x; 23 optimal.b = points[j].y-points[i].y; 24 optimal.c = points[j].x*points[i].y-points[i].x*points[j].y; 25 if( points[i].x == points[j].x && points[i].y == points[j].y ) { 26 if( Q[i][0].cnt == 1 ) { Q[i][0].cnt = Q[j][0].cnt+1; } 27 else { continue; } 28 optimal.cnt = Q[i][0].cnt; 29 for( size_t ID = 1; ID != Q[j].size(); ++ID ) { 30 Line& aLine = Q[j][ID]; 31 if( aLine.a*points[i].y + aLine.b*points[i].x + aLine.c == 0 ) { 32 if( optimal.cnt <= aLine.cnt+1 ) { optimal = aLine; ++optimal.cnt; } 33 } 34 } 35 } else { 36 optimal.cnt = 2; 37 for( size_t ID = 0; ID != Q[j].size(); ++ID ) { 38 Line& aLine = Q[j][ID]; 39 if( aLine.a*points[i].y + aLine.b*points[i].x + aLine.c == 0 ) { 40 if( optimal.cnt < aLine.cnt+1 ) { optimal.cnt = aLine.cnt+1; } 41 } 42 } 43 } 44 if( max_num < optimal.cnt ) { max_num = optimal.cnt; } 45 Q[i].push_back( optimal ); 46 } 47 } 48 delete[] Q; 49 return max_num; 50 } 51 };
Max Points on a Line,布布扣,bubuko.com
标签：style blog color 使用 io for div line
原文地址：http://www.cnblogs.com/moderate-fish/p/3901646.html
