POJ 2318 TOYS

解决玩具乱放的问题,父母通过在玩具箱内设置隔板来区分不同区域。本篇介绍了一个算法挑战,需要计算每个隔区中玩具的数量。通过输入隔板的位置和玩具落点坐标,使用二分查找及向量判断实现高效计算。

TOYS
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 17267 Accepted: 8239

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


题目大意:用几块隔板将一个箱子分成几个部分,问每个部分有多少个玩具

第一次写这类题目,用的大多是高中学的几何知识,比如这道题,如何判断一个点在线段的哪个位置,将这个点与线段的端点相连,写成向量的形式,如果两个向量相乘小于0,则点在线的左边,大于0则在线的右边,等于0则在线上

所以这道题我们就二分所有的点再判断就好了

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=5010;
int u[maxn],l[maxn];
int ans[maxn];
int n,m,x1,y1,x2,y2;
bool judge(int x,int y,int index){
	int a=(u[index]-x)*(y2-y)-(l[index]-x)*(y1-y);
	if(a<0) return false;
	else return true;
}
int main(){
	//freopen("in.txt","r",stdin);
	while(cin>>n){
		if(!n) break;
		cin>>m>>x1>>y1>>x2>>y2;
		memset(ans,0,sizeof(ans));
		for(int i=0;i<n;i++){
			cin>>u[i]>>l[i];
		}
		for(int i=0;i<m;i++){
			int x,y;
			cin>>x>>y;
			int left=0,right=n-1;
			while(left<=right){
				int mid=(left+right)>>1;
				if(judge(x,y,mid))  left=mid+1;
				else right=mid-1;
			}
			ans[left]++;
		}
		for(int i=0;i<=n;i++){
			printf("%d: %d\n",i,ans[i]);
		}
		putchar('\n');
	}
	return 0;
}

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