[POJ 2318]TOYS

[POJ 2318]TOYS

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.

题目大意

给出矩形的左上、右下坐标，按顺序给出n条线段在矩形上两条横边上的截距。这n条直线将矩形分成n+1块。给出m个点，问每个小块中有几个点。

题解：

其实一开始想用O(n^2)来做，但是考虑到这道题有多组数据，还是加了一个二分优化。

准确的来说，这道题主要用了计算几何中向量叉乘的技巧。

　　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，但可能同向也可能反向。

因此我们只需要二分查找出一个点在哪一个区间里面，然后统计一下个数就可以了。

一下是AC代码：（其实只要用叉乘，但是本人强迫症，把点乘，加，减都写进去了）

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long lol;
lol u[5001],d[5001],n,m,x1,x2,yy,y2,ans[5001];
struct Vector
{
    lol x,y;
    Vector(){}
    Vector(lol xx,lol yy){x=xx;y=yy;}
    Vector operator+(Vector b)
    {
        Vector ans;
        ans.x=x+b.x;ans.y=y+b.y;
        return ans;
    }
    Vector operator-(Vector b)
    {
        Vector ans;
        ans.x=x-b.x;ans.y=y-b.y;
        return ans;
    }
    lol operator*(Vector b)
    {
        return x*b.y-b.x*y;
    }
    lol operator^(Vector b)
    {
        return x*b.x+y*b.y;
    }
};
lol gi()
{
    lol ans=0,f=1;
    char i=getchar();
    while(i<'0'||i>'9'){if(i=='-')f=-1;i=getchar();}
    while(i>='0'&&i<='9'){ans=ans*10+i-'0';i=getchar();}
    return ans*f;
}
lol find(lol x,lol y)
{
    lol l=1,r=n,mid,ans=0;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(Vector(x-d[mid],y-y2)*Vector(u[mid]-d[mid],yy-y2)>0){ans=mid;l=mid+1;}
        else {ans=mid-1;r=mid-1;}
    }
    return ans;
}
int main()
{
    lol i;
    while(1)
    {
        memset(ans,0,sizeof(ans));
        n=gi();if(n==0)break;
        m=gi();x1=gi();yy=gi();x2=gi();y2=gi();
        for(i=1;i<=n;i++)
        {
            u[i]=gi();d[i]=gi();
        }
        for(i=1;i<=m;i++)
        {
            lol x=gi(),y=gi();
            ans[find(x,y)]++;
        }
        for(i=0;i<=n;i++)
        printf("%lld:%lld\n",i,ans[i]);
    }
    return 0;
}

 

 

 

 

转载于:https://www.cnblogs.com/huangdalaofighting/p/7263889.html