POJ 2318

二分

#include <iostream>
using namespace std;
const int maxn=5010;
int Ui[maxn],Li[maxn],dx[maxn],dy;
int amount[maxn];
int m,n,x1,y1,x2,y2;
int bsearch(int xj,int yj)
{
	int low=-1,high=m,mid;
	while(high-low>1)
	{
		mid=(high+low)/2;
		if((dy*(xj-Ui[mid])-dx[mid]*(yj-y1))<0)
			high=mid;
		else
			low=mid;
	}
	return high;
}
int main()
{
	while(scanf("%d",&m)&&m!=0)
	{
		cin>>n>>x1>>y1>>x2>>y2;
		dy=y1-y2;
		int i;
		for(i=0;i<m;i++)
		{
			cin>>Ui[i]>>Li[i];
			dx[i]=Ui[i]-Li[i];
			amount[i]=0;
		}
		amount[i]=0;
		int xj,yj,outcome;
		for(i=0;i<n;i++)
		{
			cin>>xj>>yj;
			outcome=bsearch(xj,yj);
			amount[outcome]++;
		}
		for(i=0;i<=m;i++)
		{
			cout<<i<<": "<<amount[i]<<endl;
		}
		cout<<endl;
	}
	return 0;
}


 

原文地址:https://www.cnblogs.com/lj030/p/3002317.html