《Cracking the Coding Interview》——第7章：数学和概率论——题目6

2014-03-20 02:24

题目：给定二位平面上一堆点，找到一条直线，使其穿过的点数量最多。

解法：我的解法只能适用整点，对于实数坐标就得换效率更低的办法了。请参见LeetCode - Max Points on a Line - zhuli19901106 - 博客园。

代码：

  1 // 7.6 Find the line that crosses the most points.
  2 #include <unordered_map>
  3 #include <vector>
  4 using namespace std;
  5 
  6 struct Point {
  7     int x;
  8     int y;
  9     Point() : x(0), y(0) {}
 10     Point(int a, int b) : x(a), y(b) {}
 11 };
 12 
 13 struct st {
 14     int x;
 15     int y;
 16     st(int _x = 0, int _y = 0): x(_x), y(_y) {};
 17 
 18     bool operator == (const st &other) const {
 19         return x == other.x && y == other.y;
 20     };
 21 
 22     bool operator != (const st &other) const {
 23         return x != other.x || y != other.y;
 24     };
 25 };
 26 
 27 struct line {
 28     // ax + by + c = 0;
 29     int a;
 30     int b;
 31     int c;
 32     line(int _a = 0, int _b = 0, int _c = 0): a(_a), b(_b), c(_c) {};
 33 };
 34 
 35 struct hash_functor {
 36     size_t operator () (const st &a) const {
 37         return (a.x * 1009 + a.y);
 38     }
 39 };
 40 
 41 struct compare_functor {
 42     bool operator () (const st &a, const st &b) const {
 43         return (a.x == b.x && a.y == b.y);
 44     }
 45 };
 46 
 47 typedef unordered_map<st, int, hash_functor, compare_functor> st_map;
 48 
 49 class Solution {
 50 public:
 51     int maxPoints(vector<Point> &points, line &result_line) {
 52         int n = (int)points.size();
 53         
 54         if (n <= 2) {
 55             return n;
 56         }
 57         
 58         st_map um;
 59         st tmp;
 60         // special tangent value for duplicate points
 61         st dup(0, 0);
 62         
 63         int i, j;
 64         st_map::const_iterator umit;
 65         int dup_count;
 66         int max_count;
 67         int result = 0;
 68         for (i = 0; i < n; ++i) {
 69             for (j = i + 1; j < n; ++j) {
 70                 tmp.x = points[j].x - points[i].x;
 71                 tmp.y = points[j].y - points[i].y;
 72                 // make sure one tangent value has one representation only.
 73                 normalize(tmp);
 74                 if (um.find(tmp) != um.end()) {
 75                     ++um[tmp];
 76                 } else {
 77                     um[tmp] = 1;
 78                 }
 79             }
 80             max_count = dup_count = 0;
 81             for (umit = um.begin(); umit != um.end(); ++umit) {
 82                 if (umit->first != dup) {
 83                     if (umit->second > max_count) {
 84                         max_count = umit->second;
 85                         getLine(umit->first, points[i], result_line);
 86                     }
 87                 } else {
 88                     dup_count = umit->second;
 89                 }
 90             }
 91             max_count = max_count + dup_count + 1;
 92             result = (max_count > result ? max_count : result);
 93             um.clear();
 94         }
 95         
 96         return result;
 97     }
 98 private:
 99     void normalize(st &tmp) {
100         if (tmp.x == 0 && tmp.y == 0) {
101             // (0, 0)
102             return;
103         }
104         if (tmp.x == 0) {
105             // (0, 1)
106             tmp.y = 1;
107             return;
108         }
109         if (tmp.y == 0) {
110             // (1, 0)
111             tmp.x = 1;
112             return;
113         }
114         if (tmp.x < 0) {
115             // (-2, 3) and (2, -3) => (2, -3)
116             tmp.x = -tmp.x;
117             tmp.y = -tmp.y;
118         }
119         
120         int gcd_value;
121         
122         gcd_value = (tmp.y > 0 ? gcd(tmp.x, tmp.y) : gcd(tmp.x, -tmp.y));
123         // (4, -6) and (-30, 45) => (2, -3)
124         // using a double precision risks in accuracy
125         // so I did it with a pair
126         tmp.x /= gcd_value;
127         tmp.y /= gcd_value;
128     }
129     
130     int gcd(int x, int y) {
131         // used for reduction of fraction
132         return y ? gcd(y, x % y) : x;
133     }
134     
135     void getLine(st tan, Point p, line &res) {
136         res.a = tan.y;
137         res.b = -tan.x;
138         res.c = -(res.a * p.x + res.b * p.y);
139     }
140 };
141 
142 int main()
143 {
144     vector<Point> v;
145     Solution sol;
146     int i, n;
147     Point p;
148     line res;
149     
150     while (scanf("%d", &n) == 1) {
151         for (i = 0; i < n; ++i) {
152             scanf("%d%d", &p.x, &p.y);
153             v.push_back(p);
154         }
155         sol.maxPoints(v, res);
156         printf("%d %d %d\n", res.a, res.b, res.c);
157         v.clear();
158     }
159     
160     return 0;
161 }

 

 posted on 2014-03-20 02:28  zhuli19901106  阅读(228)  评论(0)  编辑  收藏  举报