poj 2318 TOYS

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15755		Accepted: 7555

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.

Source

Rocky Mountain 2003

题意：给一个矩形，将矩形分成m块，给n个点，询问每一块的点个数

第一次写计算几何，刚开始想到的是暴力求解，数据太多把自己搞懵了，看到数据范围有点大，看大牛的博客瞬间就懂了，原文链接为：blog.youkuaiyun.com/xuh723/article/details/22221229，自己还是太水了，= =

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

#define pi acos(-1.0)
#define eps 1e-10
#define pf printf
#define sf scanf
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define e tree[rt]
#define _s second
#define _f first
#define all(x) (x).begin,(x).end
#define mem(i,a) memset(i,a,sizeof i)
#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
#define mi ((l+r)>>1)
#define sqr(x) ((x)*(x))

const int inf=0x3f3f3f3f;
int x1,y1,x2,y2,m,n,x,y,ans[5005];

struct Point
{
    int x,y;
    Point(){}
    Point(int x,int y):x(x),y(y){}
    Point operator -(const Point& b)
    {
        return Point(x-b.x,y-b.y);
    }
}l[5005],u[5005];

int cross(Point a,Point b)//叉乘，判断是在某一条直线的左端还是右端，右端返回值大于0，左端小于等于0
{
    return a.x*b.y-a.y*b.x;
}

int main()
{
    while(sf("%d",&m),m)
    {
        mem(ans,0);
        sf("%d%d%d%d%d",&n,&x1,&y1,&x2,&y2);
        for0(i,m)
        {
            sf("%d%d",&x,&y);
            u[i]=Point(x,y1),l[i]=Point(y,y2);
        }
        for0(i,n)
        {
            Point p;
            sf("%d%d",&p.x,&p.y);
            int h=0,r=m,q=m;
            while(h<=r)//二分查询该点所在位置
            {
                int mid=(h+r)>>1;
                if(cross(p-l[mid],u[mid]-l[mid])<=0)
                    r=mid-1,q=mid;
                else
                    h=mid+1;
            }
            ++ans[q];
        }
        for(int i=0;i<=m;i++)
            pf("%d: %d\n",i,ans[i]);
        pf("\n");
    }
    return 0;
}