E.智乃的小球
题目链接:https://ac.nowcoder.com/acm/contest/95335/E
题意:
有n个小球,部分向左移动,部分向右移动,碰撞后交换速度。求第k次碰撞发生的时间
思路:
二分答案,当t增大,球碰撞次数增加,反之一定减小
可以把球看作穿过彼此,并认为向左移动的小球不动,向右移动的球速度翻倍
具体实现使用双指针
不清楚为什么第二个版本二分通不过))
#include<bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define pb push_back
#define endl "\n"
#pragma GCC optimize(3)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const ll llmax=LLONG_MAX;
const int maxn=1e5+5;
const int mod=1e9+7;
int n,k;
vector<int>a;
vector<int>b;
bool check(double t)
{
int cnt=0;
int p=0,q=0;
for(int i=0;i<a.size();i++){
int l=a[i],r=a[i]+t*2;
while(l>b[p]&&p<b.size())p++;
while(r>=b[q]&&q<b.size())q++;
cnt+=(q-p);
}
return cnt>=k;
}
signed main()
{
ios::sync_with_stdio(false),cin.tie(0);
cin>>n>>k;
rep(i,1,n){
int p,v;cin>>p>>v;
if(v==1){
a.pb(p);
}else b.pb(p);
}
sort(a.begin(),a.end());
sort(b.begin(),b.end());
double l=0,r=1e9+5;
while((r-l)>=1e-7){
double mid= (l+r)/2;
if(check(mid)){
r=mid-1e-7;
}else{
l=mid+1e-7;
}
}
if(r==1e9+5)cout<<"No";else{
cout<<"Yes"<<endl;
cout<<fixed<<setprecision(6)<<r;}
return 0;
}
#include<bits/stdc++.h>
#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define pb push_back
#define endl "\n"
#pragma GCC optimize(3)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const ll llmax=LLONG_MAX;
const int maxn=1e5+5;
const int mod=1e9+7;
int n,k;
vector<int>a;
vector<int>b;
int check(double t)
{
int cnt=0;
int p=0,q=0;
for(int i=0;i<a.size();i++){
int l=a[i],r=a[i]+t*2;
while(l>b[p]&&p<b.size())p++;
while(r>=b[q]&&q<b.size())q++;
cnt+=(q-p);
}
return cnt;
}
signed main()
{
ios::sync_with_stdio(false),cin.tie(0);
cin>>n>>k;
rep(i,1,n){
int p,v;cin>>p>>v;
if(v==1){
a.pb(p);
}else b.pb(p);
}
sort(a.begin(),a.end());
sort(b.begin(),b.end());
double l=0,r=1e9+5;
double res=inf;
while((r-l)>=1e-7){
double mid= (l+r)/2;
if(check(mid)==k){
res=mid;
r=mid-1e-7;
}
else if(check(mid)<k){
l=mid+1e-6;
}else{
r=mid-1e-6;
}
}
if(res==inf)cout<<"No";else{
cout<<"Yes"<<endl;
cout<<fixed<<setprecision(6)<<res;}
return 0;
}

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