就是有n组,每组的数量是num,只能增加数字,增加的代价为t,求能使所有组的数量都不同的最小代价。

#include<bits/stdc++.h>
#define N 200005
#define endl '\n'
#define _for(i,a,b) for(int i=a;i<b;i++)
using namespace std;
typedef long long ll;
struct Node{
int num; ll t;
bool operator < (const Node &o){
return t>o.t;
}
}a[N];
map<int,int > F;
int find(int k){
if(F[k]==){
return k;
}
else return F[k]=find(F[k]);
}
void link(int a,int b){
int fa=find(a),fb=find(b);
if(fa!=fb) F[fa]=fb;
}
ll res;
int main(){
ios::sync_with_stdio(),cin.tie(),cout.tie();
int n ; cin>>n;
_for(i,,n+) cin>>a[i].num;
_for(i,,n+) cin>>a[i].t;
sort(a+,a+n+);
_for(i,,n+){
int tem=a[i].num;
int to= find(tem);
link(to,to+);
if( to == tem ) ;
else {
res+= 1ll*(to-tem)*a[i].t;
}
}
cout<<res<<endl;
return ;
}

#include<bits/stdc++.h> #define N 200005#define endl '\n' #define _for(i,a,b) for(int i=a;i<b;i++)using namespace std;typedef long long ll;  struct Node{int num; ll t;bool operator < (const Node &o){ return t>o.t;}}a[N];map<int,int > F;int find(int k){if(F[k]==0){return k;}else return F[k]=find(F[k]); }void link(int a,int b){int fa=find(a),fb=find(b);if(fa!=fb) F[fa]=fb;}ll res;int main(){    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);     int n ; cin>>n;_for(i,1,n+1) cin>>a[i].num;_for(i,1,n+1) cin>>a[i].t;sort(a+1,a+n+1);_for(i,1,n+1){int tem=a[i].num;int to= find(tem);link(to,to+1);if( to == tem ) ;else { res+= 1ll*(to-tem)*a[i].t;}}cout<<res<<endl;return 0;}

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