poj 2318 TOYS (計算幾何)

B - TOYS

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

Calculate the number of toys that land in each bin of a partitioned toy box.
Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John 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 John to find his favorite toys.

John's parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.

For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.

Input

The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). 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 contain two integers per line, Ui Li, indicating that the ends of the i-thcardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.

Output

The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.

Sample Input

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

Sample Output

0: 2
1: 1
2: 1
3: 1
4: 0
5: 1

0: 2
1: 2
2: 2
3: 2
4: 2

Hint

As the example illustrates, toys that fall on the boundary of the box are "in" the box.

題意：有一個大箱子，由n個板分爲n+1塊，標號爲0~n已知盒子左上角和右下角的座標及每個板上下

　　　兩端的橫座標（板不會交錯，且按順序給出）然後給出玩具的座標，統計每塊空間內玩具個數

　　　（保證玩具一定落在空間內）。

題解：用二分法和叉積判斷該點的位置。叉積：如果x1y2-x2y1等於0則線段12共線，如果差大於0，那麼線段1在2的右邊，

　　　否則在左邊。用數組保存一下輸出即可。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
struct point{
    int x,y;
};
struct Line{
    point a,b;
}line[5005];
int cnt[5005];
int Multi(point p1,point p2,point p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
void s(point a,int n)
{
    int l=0,r=n-1,mid;
    while(l<r)
    {
        mid=(l+r)>>1;
        if(Multi(a,line[mid].a,line[mid].b)>0) l=mid+1;
        else r=mid;
    }
    if(Multi(a,line[l].a,line[l].b)<0)
        cnt[l]++;
    else
        cnt[l+1]++;
}
int main()
{
    int n,m,x1,y1,x2,y2;
    int i,t1,t2;
    point a;
    while(cin>>n&&n)
    {
        cin>>m>>x1>>y1>>x2>>y2;
        for(int i=0;i<n;i++)
        {
            cin>>t1>>t2;
            line[i].a.x=t1;
            line[i].a.y=y1;
            line[i].b.x=t2;
            line[i].b.y=y2;
        }
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<m;i++)
        {
            cin>>a.x>>a.y;
            s(a,n);
        }
        for(int i=0;i<=n;i++)
        cout<<i<<": "<<cnt[i]<<endl;
        cout<<endl;
    }
    return 0;
}