poj 2318 叉积的应用+二分

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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.

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

0: 2
1: 1
2: 1
3: 1
4: 0
5: 1

0: 2
1: 2
2: 2
3: 2
4: 2

通过结果的正负判断两矢量之间的顺逆时针关系
若 a x b > 0表示a在b的顺时针方向上
若 a x b < 0表示a在b的逆时针方向上
若 a x b == 0表示a在b共线，但不确定方向是否相同

记玩具在点p0，某块板的上边点是p1，下边点是p2，p2p1(向量)×p2p0>0表示p0在p1p2的左面，<0表示在右面。接下来就是用二分法找出每个点所在的分区。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<string>
#include<algorithm>
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
int n,m,xx,yy,x2,y2;
struct Line
{
    int u,l;
}line[5010];
int toy[5010];
bool isplace(int x,int y,int u,int l)
{
    if((x-l)*(yy-y2)-(y-y2)*(u-l)<0)
        return true;//左
    return false;
}
int binarysearch(int x,int y)
{
    int l=0,r=n;
    while(l<r)
    {
        int mid=(l+r)/2;
        if(isplace(x,y,line[mid].u,line[mid].l))
            r=mid;
        else
            l=mid+1;
    }
    return l;
}
int main()
{
    bool flag=0;
    while(cin>>n)
    {
        if(n==0)
            return 0;
        if(flag)
            cout<<endl;
        else
            flag=1;
        scanf("%d%d%d%d%d",&m,&xx,&yy,&x2,&y2);
        for(int i=0;i<=n-1;i++)
            scanf("%d%d",&line[i].u,&line[i].l);
        memset(toy,0,sizeof(toy));
        for(int i=0;i<=m-1;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            toy[binarysearch(a,b)]++;
        }
        for(int i=0;i<=n;i++)
            cout<<i<<": "<<toy[i]<<endl;
    }
    return 0;
}

poj 2318 叉积的应用+二分

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原文地址：http://blog.csdn.net/winycg/article/details/51340837