几何+点与线段的位置关系+二分（POJ2318）

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10666		Accepted: 5128

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-th cardboard 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.

题意：给出一个长方形槽，然后从左到右给出n个隔板，把该长方形槽分成n+1个空间，n个隔板互不交叉，最后给出m个小球，随机给出，保证每个小球不在外部和隔板上；问每个空间有多少个小球？（如果不是按照顺序给出的，按照x排序即可，因为互补交叉）

分析;因为隔板是按照顺序给出的，那么对于每个小球二分隔板，用叉乘判断点在线段的位置，枚举答案即可：

#include"string.h"
#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"queue"
#include"stack"
#define M 5009
#define N 100009
#include"stdlib.h"
#include"math.h"
#define inf 10000000000000000LL
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define eps 1e-10
using namespace std;
struct node
{
    double x,y;
    node(){}
    node(double _x,double _y):x(_x),y(_y){}
    node operator +(node p)
    {
        return node(x+p.x,y+p.y);
    }
    node operator -(node p)
    {
        return node(x-p.x,y-p.y);
    }
    double operator *(node p)
    {
        return x*p.y-y*p.x;
    }
    double operator ^(node p)
    {
        return x*p.x+y*p.y;
    }
}p[M],k[M];
int n;
int cmp(node a,node b)
{
    return a.x<b.x;
}
double len(node a)
{
    return sqrt(a^a);
}
double dis(node a,node b)
{
    return len(b-a);
}
double cross(node a,node b,node c)
{
    return (b-a)*(c-a);
}
double dot(node a,node b,node c)
{
    return (b-a)^(c-a);
}
int fun(node p,node *q)
{
    double rad=0;
    for(int i=1;i<=4;i++)
    {
        rad+=acos(dot(p,q[i-1],q[i%4])/dis(p,q[i-1])/dis(p,q[i%4]));
    }
    if(fabs(rad-PI*2)<eps)
        return 1;
    else
        return 0;
}
int s[M],vis[M];
int main()
{
    double x1,x2,y1,y2;
    int n,m,i,kk=0;
    while(scanf("%d",&n),n)
    {
        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        k[n+1]=node(x2,x2);
        for(i=1;i<=n;i++)
            scanf("%lf%lf",&k[i].x,&k[i].y);
        memset(s,0,sizeof(s));
        for(i=1;i<=m;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
            int l=1,r=n+1,mid;
            int fuck;
            while(l<=r)
            {
                mid=(l+r)/2;
                node p1(k[mid].y,y2);
                node p2(k[mid].x,y1);
                if(cross(p1,p2,p[i])>0)
                {
                    fuck=mid;
                    r=mid-1;
                }
                else
                {
                    l=mid+1;
                }

            }
            s[fuck-1]++;
        }
        if(kk)
            printf("\n");
        kk++;
        for(i=0;i<=n;i++)
            printf("%d: %d\n",i,s[i]);
    }
}