POJ 2318 TOYS (判断点与直线关系+二分查找)

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12533		Accepted: 6066

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：一个盒子，中间又分好多格子，给你中间的隔板的起点和终点坐标，然后又有m个玩具，给你玩具的坐标，然后求每个格子有多少玩具

思路：要判断玩具在那个格子，需寻找一条在其上面的直线，即此点在直线的下面，根据直线起点终点求出方程，然后代入坐标，如果小于0说明在直线下面，然后二分查找这条直线，最后输出就行了（ps：输出要包含最后一个格子）
ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 60001
#define LL long long
#define INF 0xfffffff
#define fab(x) (x)>0?(x):(-x)
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
struct s
{
	int a,b,c,d;
}line[MAXN];
int ans[MAXN];
int n,m;
int check(int x,int y,int i)//检查和直线的关系，返回代入的方程值
{
	double ans=(line[i].b-line[i].d)*x+(line[i].c-line[i].a)*y+(line[i].a*line[i].d-line[i].c*line[i].b);
    if(ans<0)
    return 1;
    return 0;
}
void solve(int x,int y)
{
	int low=0;
	int high=n-1;
	int mid;
	while(low<=high)//二分查找
	{
		mid=(low+high)/2;
		if(check(x,y,mid)==0)
		low=mid+1;
		else
		high=mid-1;
	}
	ans[low]++;
}
int main()
{
	int t,i,x1,y1,x2,y2,aa,bb;
	while(scanf("%d",&n)!=EOF,n)
	{
		scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&aa,&bb);
			line[i].a=aa;
			line[i].b=y1;
			line[i].c=bb;
			line[i].d=y2;
		}
		mem(ans);
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&aa,&bb);
			solve(aa,bb);
		}
		for(i=0;i<=n;i++)
		printf("%d: %d\n",i,ans[i]);
		printf("\n");
	}
	return 0;
}