容易发现答案只与w1和w2的比值有关，而与具体数值无关
那么先特殊算一下w1=0和w2等于0的情况，然后就直接假设w1=1
然后的话对于每个物品可能有4种情况：永远比1号优，永远比1号劣，永远和1号相等，当w2<某值时比1号优，等于时相等，否则比一号劣，或者当w2>某值时比1号劣，等于时相等，否则比1号优
对于每个手机算一下他的情况和w2的分界点，然后按分界点排序之后扫一遍即可
比赛的时候有个地方x和y打反一直wa，然后就弃疗了
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<vector>
#include<set>
using namespace std;
#define MAXN 100010
#define MAXM 1010
#define INF 1000000000
#define MOD 1000000007
#define ll long long
#define eps 1e-8
struct data{
double v;
int c;
friend bool operator <(data x,data y){
return x.v<y.v;
}
};
int n;
int x[MAXN],y[MAXN];
int low=1,high=INF;
data t[MAXN];
int tot;
int main(){
int i;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d%d",&x[i],&y[i]);
}
int tlowx=1,thighx=1,tlowy=1,thighy=1;
for(i=2;i<=n;i++){
if(x[i]>x[1]){
thighx++;
}
if(x[i]>=x[1]){
tlowx++;
}
if(y[i]>y[1]){
thighy++;
}
if(y[i]>=y[1]){
tlowy++;
}
}
low=max(low,max(tlowx,tlowy));
high=min(high,min(thighx,thighy));
int tlow=1,thigh=1;
for(i=2;i<=n;i++){
if(y[i]==y[1]){
if(x[i]>=x[1]){
tlow++;
}
if(x[i]>x[1]){
thigh++;
}
continue ;
}
if(y[i]<y[1]){
t[++tot].v=1000.*(x[1]-x[i])/(y[i]-y[1]);
t[tot].c=-1;
tlow++;
thigh++;
}
if(y[i]>y[1]){
t[++tot].v=1000.*(x[1]-x[i])/(y[i]-y[1]);
t[tot].c=1;
}
if(t[tot].v<=0){
tlow+=t[tot].c;
thigh+=t[tot].c;
tot--;
}
}
if(tot){
sort(t+1,t+tot+1);
for(i=1;i<=tot;){
int wzh=i;
int tc0=0,tc1=0;
while(wzh<=tot&&fabs(t[wzh].v-t[i].v)<eps){
if(t[wzh].c==1){
tc1++;
}else{
tc0++;
}
wzh++;
}
low=max(low,tlow+tc1);
high=min(high,thigh-tc0);
tlow+=tc1-tc0;
thigh+=tc1-tc0;
i=wzh;
}
}
printf("%d %d\n",high,low);
return 0;
}
/*
2
3 2
1 2
*/