题目传送门

题意：POJ 2318 有一个长方形，用线段划分若干区域，给若干个点，问每个区域点的分布情况

分析：点和线段的位置判断可以用叉积判断。给的线段是排好序的，但是点是无序的，所以可以用二分优化。用到了叉积

  1. /************************************************
  2. * Author        :Running_Time
  3. * Created Time  :2015/10/23 星期五 11:38:18
  4. * File Name     :POJ_2318.cpp
  5.  ************************************************/
  6.  
  7. #include <cstdio>
  8. #include <algorithm>
  9. #include <iostream>
  10. #include <sstream>
  11. #include <cstring>
  12. #include <cmath>
  13. #include <string>
  14. #include <vector>
  15. #include <queue>
  16. #include <deque>
  17. #include <stack>
  18. #include <list>
  19. #include <map>
  20. #include <set>
  21. #include <bitset>
  22. #include <cstdlib>
  23. #include <ctime>
  24. using namespace std;
  25.  
  26. #define lson l, mid, rt << 1
  27. #define rson mid + 1, r, rt << 1 | 1
  28. typedef long long ll;
  29. const int N = 5e3 + 10;
  30. const int INF = 0x3f3f3f3f;
  31. const int MOD = 1e9 + 7;
  32. const double EPS = 1e-10;
  33. struct Point    {       //点的定义
  34.     double x, y;
  35.     Point (double x=0, double y=0) : x (x), y (y) {}
  36. };
  37. typedef Point Vector;       //向量的定义
  38. Point read_point(void)   {      //点的读入
  39.     double x, y;
  40.     scanf ("%lf%lf", &x, &y);
  41.     return Point (x, y);
  42. }
  43. double polar_angle(Vector A)  {     //向量极角
  44.     return atan2 (A.y, A.x);
  45. }
  46. double dot(Vector A, Vector B)  {       //向量点积
  47.     return A.x * B.x + A.y * B.y;
  48. }
  49. double cross(Vector A, Vector B)    {       //向量叉积
  50.     return A.x * B.y - A.y * B.x;
  51. }
  52. int dcmp(double x)  {       //三态函数，减少精度问题
  53.     if (fabs (x) < EPS) return 0;
  54.     else    return x < 0 ? -1 : 1;
  55. }
  56. Vector operator + (Vector A, Vector B)  {       //向量加法
  57.     return Vector (A.x + B.x, A.y + B.y);
  58. }
  59. Vector operator - (Vector A, Vector B)  {       //向量减法
  60.     return Vector (A.x - B.x, A.y - B.y);
  61. }
  62. Vector operator * (Vector A, double p)  {       //向量乘以标量
  63.     return Vector (A.x * p, A.y * p);
  64. }
  65. Vector operator / (Vector A, double p)  {       //向量除以标量
  66.     return Vector (A.x / p, A.y / p);
  67. }
  68. bool operator < (const Point &a, const Point &b)    {       //点的坐标排序
  69.     return a.x < b.x || (a.x == b.x && a.y < b.y);
  70. }
  71. bool operator == (const Point &a, const Point &b)   {       //判断同一个点
  72.     return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
  73. }
  74. double length(Vector A) {       //向量长度，点积
  75.     return sqrt (dot (A, A));
  76. }
  77. double angle(Vector A, Vector B)    {       //向量转角，逆时针，点积
  78.     return acos (dot (A, B) / length (A) / length (B));
  79. }
  80. double area_triangle(Point a, Point b, Point c) {       //三角形面积，叉积
  81.     return fabs (cross (b - a, c - a)) / 2.0;
  82. }
  83. Vector rotate(Vector A, double rad) {       //向量旋转，逆时针
  84.     return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
  85. }
  86. Vector nomal(Vector A)  {       //向量的单位法向量
  87.     double len = length (A);
  88.     return Vector (-A.y / len, A.x / len);
  89. }
  90. Point point_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点，参数方程
  91.     Vector U = p - q;
  92.     double t = cross (W, U) / cross (V, W);
  93.     return p + V * t;
  94. }
  95. double dis_to_line(Point p, Point a, Point b)   {       //点到直线的距离，两点式
  96.     Vector V1 = b - a, V2 = p - a;
  97.     return fabs (cross (V1, V2)) / length (V1);
  98. }
  99. double dis_to_seg(Point p, Point a, Point b)    {       //点到线段的距离，两点式
  100.  
  101.     if (a == b) return length (p - a);
  102.     Vector V1 = b - a, V2 = p - a, V3 = p - b;
  103.     if (dcmp (dot (V1, V2)) < 0)    return length (V2);
  104.     else if (dcmp (dot (V1, V3)) > 0)   return length (V3);
  105.     else    return fabs (cross (V1, V2)) / length (V1);
  106. }
  107. Point point_proj(Point p, Point a, Point b)   {     //点在直线上的投影，两点式
  108.     Vector V = b - a;
  109.     return a + V * (dot (V, p - a) / dot (V, V));
  110. }
  111. bool inter(Point a1, Point a2, Point b1, Point b2)  {       //判断线段相交，两点式
  112.     double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
  113.            c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
  114.     return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
  115. }
  116. bool on_seg(Point p, Point a1, Point a2)    {       //判断点在线段上，两点式
  117.     return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
  118. }
  119. double area_poly(Point *p, int n)   {       //多边形面积
  120.     double ret = 0;
  121.     for (int i=1; i<n-1; ++i)   {
  122.         ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
  123.     }
  124.     return ret / 2;
  125. }
  126. struct Line {
  127.     Point s, e;
  128.     Line () {}
  129.     Line (Point s, Point e) : s (s), e (e) {}
  130. };
  131. Line board[N];
  132. Point toy[N];
  133. Point b, e;
  134. int ans[N];
  135.  
  136. int main(void)    {
  137.     int n, m;
  138.     bool fir = true;
  139.     while(scanf ("%d", &n) == 1)    {
  140.         if (!n) break;
  141.         if (fir)    fir = false;
  142.         else    puts ("");
  143.         scanf ("%d", &m);
  144.         b = read_point ();
  145.         e = read_point ();
  146.         double U, L;
  147.         for (int i=0; i<n; ++i) {
  148.             scanf ("%lf%lf", &U, &L);
  149.             board[i] = Line (Point (U, b.y), Point (L, e.y));
  150.         }
  151.         board[n] = Line (Point (e.x, b.y), Point (e.x, e.y));
  152.         for (int i=0; i<m; ++i) {
  153.             toy[i] = read_point ();
  154.         }
  155.         memset (ans, 0, sizeof (ans));
  156.         for (int i=0; i<m; ++i) {
  157.             if (toy[i].x < b.x || toy[i].x > e.x || toy[i].y > b.y || toy[i].y < e.y)   continue;
  158.             int l = 0, r = n, mid, tmp = 0;
  159.             while (l <= r)   {
  160.                 mid = (l + r) >> 1;
  161.                 if (cross (board[mid].s - toy[i], board[mid].e - toy[i]) < 0)   {
  162.                     tmp = mid;
  163.                     r = mid - 1;
  164.                 }
  165.                 else    l = mid + 1;
  166.             }
  167.             ans[tmp]++;
  168.         }
  169.         for (int i=0; i<=n; ++i)    {
  170.             printf ("%d: %d\n", i, ans[i]);
  171.         }
  172.     }
  173.  
  174.     return 0;
  175. }

题目传送门

题意：POJ 2398 和上面那题没什么区别，回答的问题不同，线段无序先排序。

  1. /************************************************
  2. * Author        :Running_Time
  3. * Created Time  :2015/10/23 星期五 11:38:18
  4. * File Name     :POJ_2398.cpp
  5.  ************************************************/
  6.  
  7. #include <cstdio>
  8. #include <algorithm>
  9. #include <iostream>
  10. #include <sstream>
  11. #include <cstring>
  12. #include <cmath>
  13. #include <string>
  14. #include <vector>
  15. #include <queue>
  16. #include <deque>
  17. #include <stack>
  18. #include <list>
  19. #include <map>
  20. #include <set>
  21. #include <bitset>
  22. #include <cstdlib>
  23. #include <ctime>
  24. using namespace std;
  25.  
  26. #define lson l, mid, rt << 1
  27. #define rson mid + 1, r, rt << 1 | 1
  28. typedef long long ll;
  29. const int N = 1e3 + 10;
  30. const int INF = 0x3f3f3f3f;
  31. const int MOD = 1e9 + 7;
  32. const double EPS = 1e-10;
  33. struct Point    {       //点的定义
  34.     double x, y;
  35.     Point (double x=0, double y=0) : x (x), y (y) {}
  36. };
  37. typedef Point Vector;       //向量的定义
  38. Point read_point(void)   {      //点的读入
  39.     double x, y;
  40.     scanf ("%lf%lf", &x, &y);
  41.     return Point (x, y);
  42. }
  43. double polar_angle(Vector A)  {     //向量极角
  44.     return atan2 (A.y, A.x);
  45. }
  46. double dot(Vector A, Vector B)  {       //向量点积
  47.     return A.x * B.x + A.y * B.y;
  48. }
  49. double cross(Vector A, Vector B)    {       //向量叉积
  50.     return A.x * B.y - A.y * B.x;
  51. }
  52. int dcmp(double x)  {       //三态函数，减少精度问题
  53.     if (fabs (x) < EPS) return 0;
  54.     else    return x < 0 ? -1 : 1;
  55. }
  56. Vector operator + (Vector A, Vector B)  {       //向量加法
  57.     return Vector (A.x + B.x, A.y + B.y);
  58. }
  59. Vector operator - (Vector A, Vector B)  {       //向量减法
  60.     return Vector (A.x - B.x, A.y - B.y);
  61. }
  62. Vector operator * (Vector A, double p)  {       //向量乘以标量
  63.     return Vector (A.x * p, A.y * p);
  64. }
  65. Vector operator / (Vector A, double p)  {       //向量除以标量
  66.     return Vector (A.x / p, A.y / p);
  67. }
  68. bool operator < (const Point &a, const Point &b)    {       //点的坐标排序
  69.     return a.x < b.x || (a.x == b.x && a.y < b.y);
  70. }
  71. bool operator == (const Point &a, const Point &b)   {       //判断同一个点
  72.     return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
  73. }
  74. double length(Vector A) {       //向量长度，点积
  75.     return sqrt (dot (A, A));
  76. }
  77. double angle(Vector A, Vector B)    {       //向量转角，逆时针，点积
  78.     return acos (dot (A, B) / length (A) / length (B));
  79. }
  80. double area_triangle(Point a, Point b, Point c) {       //三角形面积，叉积
  81.     return fabs (cross (b - a, c - a)) / 2.0;
  82. }
  83. Vector rotate(Vector A, double rad) {       //向量旋转，逆时针
  84.     return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
  85. }
  86. Vector nomal(Vector A)  {       //向量的单位法向量
  87.     double len = length (A);
  88.     return Vector (-A.y / len, A.x / len);
  89. }
  90. Point point_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点，参数方程
  91.     Vector U = p - q;
  92.     double t = cross (W, U) / cross (V, W);
  93.     return p + V * t;
  94. }
  95. double dis_to_line(Point p, Point a, Point b)   {       //点到直线的距离，两点式
  96.     Vector V1 = b - a, V2 = p - a;
  97.     return fabs (cross (V1, V2)) / length (V1);
  98. }
  99. double dis_to_seg(Point p, Point a, Point b)    {       //点到线段的距离，两点式
  100.  
  101.     if (a == b) return length (p - a);
  102.     Vector V1 = b - a, V2 = p - a, V3 = p - b;
  103.     if (dcmp (dot (V1, V2)) < 0)    return length (V2);
  104.     else if (dcmp (dot (V1, V3)) > 0)   return length (V3);
  105.     else    return fabs (cross (V1, V2)) / length (V1);
  106. }
  107. Point point_proj(Point p, Point a, Point b)   {     //点在直线上的投影，两点式
  108.     Vector V = b - a;
  109.     return a + V * (dot (V, p - a) / dot (V, V));
  110. }
  111. bool inter(Point a1, Point a2, Point b1, Point b2)  {       //判断线段相交，两点式
  112.     double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
  113.            c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
  114.     return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
  115. }
  116. bool on_seg(Point p, Point a1, Point a2)    {       //判断点在线段上，两点式
  117.     return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
  118. }
  119. double area_poly(Point *p, int n)   {       //多边形面积
  120.     double ret = 0;
  121.     for (int i=1; i<n-1; ++i)   {
  122.         ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
  123.     }
  124.     return ret / 2;
  125. }
  126. struct Line {
  127.     Point s, e;
  128.     Line () {}
  129.     Line (Point s, Point e) : s (s), e (e) {}
  130. };
  131. bool cmp(Line A, Line B)    {
  132.     return A.s.x < B.s.x || (A.s.x == B.s.x && A.e.x < B.e.x);
  133. }
  134. Line board[N];
  135. Point toy[N];
  136. Point b, e;
  137. int ans[N];
  138.  
  139. int main(void)    {
  140.     int n, m;
  141.     while(scanf ("%d", &n) == 1)    {
  142.         if (!n) break;
  143.         scanf ("%d", &m);
  144.         b = read_point ();
  145.         e = read_point ();
  146.         double U, L;
  147.         for (int i=0; i<n; ++i) {
  148.             scanf ("%lf%lf", &U, &L);
  149.             board[i] = Line (Point (U, b.y), Point (L, e.y));
  150.         }
  151.         board[n] = Line (Point (e.x, b.y), Point (e.x, e.y));
  152.         sort (board, board+1+n, cmp);
  153.         for (int i=0; i<m; ++i) {
  154.             toy[i] = read_point ();
  155.         }
  156.         memset (ans, 0, sizeof (ans));
  157.         for (int i=0; i<m; ++i) {
  158.             if (toy[i].x < b.x || toy[i].x > e.x || toy[i].y > b.y || toy[i].y < e.y)   continue;
  159.             int l = 0, r = n, mid, tmp = 0;
  160.             while (l <= r)   {
  161.                 mid = (l + r) >> 1;
  162.                 if (cross (board[mid].s - toy[i], board[mid].e - toy[i]) < 0)   {
  163.                     tmp = mid;
  164.                     r = mid - 1;
  165.                 }
  166.                 else    l = mid + 1;
  167.             }
  168.             ans[tmp]++;
  169.         }
  170.         puts ("Box");
  171.         for (int i=1; i<=m; ++i)    {
  172.             int cnt = 0;
  173.             for (int j=0; j<=n; ++j)    {
  174.                 if (ans[j] == i)    ++cnt;
  175.             }
  176.             if (cnt)    {
  177.                 printf ("%d: %d\n", i, cnt);
  178.             }
  179.         }
  180.     }
  181.  
  182.     return 0;
  183. }

　　

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