POJ-2398

Toy Storage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4243		Accepted: 2517

Description

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard
partitions into the box. Even if Reza keeps throwing his toys into the
box, at least toys that get thrown into different partitions stay
separate. The box looks like this from the top:



We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.

Input

The
input consists of a number of cases. The first line consists of six
integers n, m, x1, y1, x2, y2. The number of cardboards to form the
partitions is n (0 < n <= 1000) and the number of toys is given in
m (0 < m <= 1000). The coordinates of the upper-left corner and
the lower-right corner of the box are (x1, y1) and (x2, y2),
respectively. The following n lines each consists of two integers Ui Li,
indicating that the ends of the ith cardboard is at the coordinates
(Ui, y1) and (Li, y2). You may assume that the cardboards do not
intersect with each other. The next m lines each consists of two
integers Xi Yi specifying where the ith toy has landed in the box. You
may assume that no toy will land on a cardboard.

A line consisting of a single 0 terminates the input.

Output

For
each box, first provide a header stating "Box" on a line of its own.
After that, there will be one line of output per count (t > 0) of
toys in a partition. The value t will be followed by a colon and a
space, followed the number of partitions containing t toys. Output will
be sorted in ascending order of t for each box.

Sample Input

4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0

Sample Output

Box
2: 5
Box
1: 4
2: 1

Source

Tehran 2003 Preliminary

/**
          题意：给一个区域的左上和右上坐标，然后给出区域中的n条边，然后，给出m个玩具的坐标；
                    但是没有排序，要求比如说是有一个玩具的区域的个数
          做法：计算几何
**/
#include <iostream>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<stdio.h>
#define maxn 5000 + 10
using namespace std;
int n,m;
int mmap[maxn];
int mmpp[maxn];
const double eps = 1e-;
const  double PI = acos(-1.0);
struct Point
{
          double x;
          double y;
          Point(){}
          Point(double _x,double _y)
          {
                    x = _x;
                    y = _y;
          }
          Point operator - (const Point &b) const
          {
                    return Point(x - b.x,y - b.y);
          }
          double operator ^(const Point &b) const
          {
                    return x*b.y - y*b.x;
          }
          double  operator *(const Point &b) const
          {
                    return x*b.x + y*b.y;
          }
}point[maxn];
struct Line
{
          Point s;
          Point e;
          Line(){}
          Line(Point _s,Point _e)
          {
                    s = _s;
                    e = _e;
          }
}line[maxn];
bool  cmp(Line a,Line b)
{
          return a.s.x < b.s.x;
}
int mul(Point a,Point b,Point c)
{
          return (b - a) ^(c - a);
}
int main()
{
#ifndef ONLINE_JUDGE
          freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
          int x1,y1,x2,y2;
          while(~scanf("%d",&n))
          {
                    if(n == ) break;
                    memset(mmap,,sizeof(mmap));
                    memset(mmpp,,sizeof(mmpp));
                    scanf("%d %d %d %d %d",&m,&x1,&y1,&x2,&y2);
                    int u,l;
                    for(int i=;i<n;i++)
                    {
                              scanf("%d %d",&u,&l);
                              line[i] = Line(Point(u,y1),Point(l,y2));
                    }
                    line[n] = Line(Point(x2,y1),Point(x2,y2));
                    sort(line,line+n+,cmp);
                    int x,y;
                    for(int i=;i<m;i++)
                    {
                              scanf("%d %d",&x,&y);
                              Point pp = Point(x,y);
                              int left = ;
                              int right = n;
                              int tmp = ;
                              while(left <= right)
                              {
                                        int mid = (left + right) >>;
                                        if(mul(pp,line[mid].s,line[mid].e)<)
                                        {
                                                  tmp = mid;
                                                  right = mid-;
                                        }
                                        else left = mid+;
                              }
                              mmap[tmp] ++;
                    }
                    int mmax = ;
                    for(int i=;i<=n;i++)
                    {
                              mmpp[i] = ;
                    }
                    for(int i=;i<=n;i++)
                    {
                             if(mmap[i]>) mmpp[mmap[i]] ++;
                    }
                    printf("Box\n");
                    for(int i=;i<=n;i++)
                    {
                              if(mmpp[i]>) printf("%d: %d\n",i,mmpp[i]);
                    }
          }
          return ;
}