https://codeforces.com/contest/1140/problem/C
题意
每首歌有\(t_i\)和\(b_i\)两个值,最多挑选m首歌,使得sum(\(t_i\))*min(\(b_i\))最大
题解
- 假如最小的\(b_i\)确定了,那么拿得越多越好
- 枚举最小的\(b_i\)然后取剩下最大的\(t_i\)
- 两种做法
1.从\(b_i\)大向小扫,这样用优先队列维护最大的那些\(t_i\)(丢弃最小的)
2.用两个优先队列模拟
代码
//做法1
#include<bits/stdc++.h>
#define ll long long
#define pii pair<ll,ll>
#define ft first
#define se second
using namespace std;
priority_queue<ll,vector<ll>,greater<ll> >Q;
pii p[300005];
ll ans,sum;
int n,k;
int main(){
cin>>n>>k;
for(int i=0;i<n;i++)scanf("%lld%lld",&p[i].se,&p[i].ft);
sort(p,p+n,greater<pii>());
for(int i=0;i<k;i++){
Q.push(p[i].se);
sum+=p[i].se;
ans=max(ans,sum*p[i].ft);
}
for(int i=k;i<n;i++){
ll mi=Q.top();
if(p[i].se<=mi)continue;
else{
Q.pop();
Q.push(p[i].se);
sum+=p[i].se-mi;
ans=max(ans,sum*p[i].ft);
}
}
cout<<ans;
}//做法2
#include<bits/stdc++.h>
#define ll long long
#define pii pair<ll,ll>
#define mk make_pair
#define ft first
#define se second
using namespace std;
struct N{
ll t,d;
}p[300005];
ll n,k,ans,i,tim;
struct cmp{
bool operator() (pii a, pii b ){
if(a.ft==b.ft)return a.se>b.se;
return a.ft>b.ft;
}
};
bool cmp1(N x,N y){
return x.d<y.d;
}
priority_queue<pii,vector<pii>,cmp> s;
priority_queue<pii>q;
int main(){
cin>>n>>k;
for(i=1;i<=n;i++){
scanf("%lld%lld",&p[i].t,&p[i].d);
q.push(mk(p[i].t,p[i].d));
}
sort(p+1,p+n+1,cmp1);
tim=0;
for(i=1;i<=n;i++){
while(!s.empty()&&s.top().ft<p[i].d){
//cout<<"s "<<s.top().se<<endl;
tim-=s.top().se;s.pop();
}
ans=max(ans,p[i].d*tim);
while(!q.empty()&&s.size()<k){
//cout<<"q "<<q.top().ft<<endl;
if(q.top().se>=p[i].d){
tim+=q.top().ft;
s.push(mk(q.top().se,q.top().ft));
}
q.pop();
ans=max(ans,p[i].d*tim);
}
}
cout<<ans;
}