鏈接：https://cn.vjudge.net/problem/POJ-2398
題解：只需要排個序，其餘和https://blog.csdn.net/birdmanqin/article/details/100000498 相同。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;
const int N = 1e3+10;
struct node
{
double ux,dx;
}a[N];
struct seg
{
double k,b;
bool flag;
}segs[N];
int ans[N];
int res[N];
double ux,dx,x,y,x1,y1,x2,y2;
int n,m;
bool cmp(node a,node b)
{
if(a.ux==b.ux)
return a.dx<b.dx;
else return a.ux<b.ux;
}
void Init()
{
memset(ans,0,sizeof(ans));
memset(res,0,sizeof(res));
}
int judge(double x,double y)
{
for(int i=1;i<=n;i++)
{
if(segs[i].flag)
{
if(segs[i].k>x)
return i-1;
else continue;
}
if((y-segs[i].b)/segs[i].k>x)
return i-1;
}
return n;
}
int main(void)
{
while(~scanf("%d",&n)&&n)
{
Init();
scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
for(int i=1;i<=n;i++)
scanf("%lf%lf",&a[i].ux,&a[i].dx);
sort(a+1,a+1+n,cmp);
for(int i=1;i<=n;i++)
{
ux=a[i].ux,dx=a[i].dx;
if(ux==dx)
{
segs[i].flag=1;
segs[i].k=ux;
continue;
}
segs[i].flag=0;
segs[i].k=(y1-y2)/(ux-dx);
segs[i].b=y1-segs[i].k*ux;
}
for(int i=1;i<=m;i++)
{
scanf("%lf%lf",&x,&y);
ans[judge(x,y)]++;
}
for(int i=0;i<=n;i++)
if(ans[i])
res[ans[i]]++;
puts("Box");
for(int i=1;i<=m;i++)
{
if(res[i])
{
printf("%d: %d",i,res[i]);
putchar('\n');
}
}
}
return 0;
}