poj 2318 叉积+二分

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13262		Accepted: 6412

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.

/*
poj 2318 叉积+二分

一个矩形，有被若干直线分成N个格子，给出一个点的坐标，问你该点位于哪个点中。
知识点：其实就是点在凸四边形内的判断，若利用叉积的性质，可以二分求解。

叉积的结果也是一个向量，是垂直于向量a，b所形成的平面，如果看成三维坐标的话是在 z 轴上，上面结果是它的模。
方向判定：右手定则，（右手半握，大拇指垂直向上，四指右向量a握向b，大拇指的方向就是叉积的方向）
叉积的意义：
1：其结果是a和b为相邻边形成平行四边形的面积。
2：结果有正有负，有sin（a，b）可知和其夹角有关，夹角大于180°为负值。
本题可以通过叉积的正负来判断它在直线的哪边

hhh-2016-05-04 19:49:26
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
const int  maxn = 40010;
int tot;
int mod;
int n,m;
int x1,x2,y1,y2;

struct Point
{
    int x,y;
    Point() {}
    Point(int _x,int _y)
    {
        x = _x,y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x-b.x,y-b.y);
    }
    int operator ^(const Point &b)const
    {
        return x*b.y-y*b.x;
    }
};

struct Line
{
    Point s,t;
    Line() {}
    Line(Point _s,Point _t)
    {
        s = _s;
        t = _t;
    }
};
int tans[maxn];
Line line[maxn];
Point p;
int cal(int mid)
{
    return (line[mid].t-p)^(line[mid].s-p);
}

int main()
{
    int flag = 1;
    while(scanf("%d",&n) && n)
    {
        if(!flag)
            printf("\n");
        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
        flag = 0;
        line[n] = Line(Point(x2,y1),Point(x2,y2));
        memset(tans,0,sizeof(tans));
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d",&x1,&x2);
            line[i] = Line(Point(x1,y1),Point(x2,y2));
        }
        while(m--)
        {
            scanf("%d%d",&x1,&y1);
            int l = 0, r = n;
            int mid,ans;
            p = Point(x1,y1);
            while(l <= r)
            {
                mid = (l+r)>>1;
                if(cal(mid) > 0)
                {
                    ans = mid;
                    r = mid-1;
                }
                else
                {
                    l = mid+1;
                }
            }
            tans[ans]++;
        }
        for(int i = 0; i <= n; i++)
        {
            printf("%d: %d\n",i,tans[i]);
        }
    }
    return 0;
}