poj 2318(叉积判断点在线段的哪一侧)

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13120		Accepted: 6334

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


题意：给定n根线段将平面分成n+1个区域,然后给m个点,分散在大平面内，问每个区域内有多少个点
分析：对于每个点，我们只要利用叉积判断是否在某条线段逆时针方向就行了.这个题的点数应该开10000.
有二分的做法，比我这个应该要快不少，暴力938MS
叉积的性质：设矢量 P = (x1, y1), Q = (x2, y2)，则 P * Q = x1 * y2 - x2 * y1; 其结果是一个由 (0, 0), P, Q, P + Q 所组成的平行四边形的 带符号的面积，P * Q = -(Q * P), P * (- Q) = -(P * Q)。
           叉积的一个非常重要的性质是可以通过它的符号来判断两矢量相互之间的顺逆时针关系：
            若 P * Q > 0，则 P 在 Q 的顺时针方向；
            若 P * Q < 0, 则 P 在 Q 的逆时针方向；
            若 P * Q = 0，则 P 与 Q 共线，但不确定 P, Q 的方向是否相同；

#include <iostream>
#include <cstdio>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = 10005;
struct Point
{
    int x,y;
} p[N],q[N];
int n,m,x1,y11,x2,y2;

bool used[N];///判断点是否已经被选过了
int cnt[N]; ///判断某区域的点数量

int mult(Point a,Point b,Point c){
    return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}

int main()
{
    while(scanf("%d",&n)!=EOF,n)
    {
        scanf("%d%d%d%d%d",&m,&x1,&y11,&x2,&y2);
        memset(used,false,sizeof(used));
        memset(cnt,0,sizeof(cnt));
        int k=1;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&p[k].x,&p[k+1].x);
            p[k].y=y11,p[k+1].y=y2;
            k+=2;
        }
        for(int i=0;i<m;i++){
            scanf("%d%d",&q[i].x,&q[i].y);
        }
        int sum=0;
        for(int i=1;i<=n;i++){
            for(int j=0;j<m;j++){
                if(mult(p[2*i-1],q[j],p[2*i])>0&&!used[j]){
                    cnt[i-1]++;
                    used[j]=true;
                }
            }
            sum+=cnt[i-1];
        }
        cnt[n] = m-sum;
        for(int i=0;i<=n;i++){
            printf("%d: %d\n",i,cnt[i]);
        }
        printf("\n");
    }
    return 0;
}