排完序以後從小到大掃一遍 如果有相同的 就把其中一個+k 扔到一個數組裏 然後每次也跟這個數組比較下有沒有重的 因爲一次只能給一個ai加x最後看看要多大 加的時候如果有一本書兩個人都like 但一個人到k了 就把這個人最大的單獨like的去掉 換成這個兩個人都like的加完以後 再檢查一下後面有沒有兩個都like的 更優的
Easy and hard versions are actually different problems, so read statements of both problems completely and carefully.
Summer vacation has started so Alice and Bob want to play and joy, but… Their mom doesn’t think so. She says that they have to read some amount of books before all entertainments. Alice and Bob will read each book together to end this exercise faster.
There are n books in the family library. The i-th book is described by three integers: ti — the amount of time Alice and Bob need to spend to read it, ai (equals 1 if Alice likes the i-th book and 0 if not), and bi (equals 1 if Bob likes the i-th book and 0 if not).
So they need to choose some books from the given n books in such a way that:
Alice likes at least k books from the chosen set and Bob likes at least k books from the chosen set;
the total reading time of these books is minimized (they are children and want to play and joy as soon a possible).
The set they choose is the same for both Alice an Bob (it’s shared between them) and they read all books together, so the total reading time is the sum of ti over all books that are in the chosen set.
Your task is to help them and find any suitable set of books or determine that it is impossible to find such a set.
Input
The first line of the input contains two integers n and k (1≤k≤n≤2⋅105).
The next n lines contain descriptions of books, one description per line: the i-th line contains three integers ti, ai and bi (1≤ti≤104, 0≤ai,bi≤1), where:
ti — the amount of time required for reading the i-th book;
ai equals 1 if Alice likes the i-th book and 0 otherwise;
bi equals 1 if Bob likes the i-th book and 0 otherwise.
Output
If there is no solution, print only one integer -1. Otherwise print one integer T — the minimum total reading time of the suitable set of books.
Examples
inputCopy
8 4
7 1 1
2 1 1
4 0 1
8 1 1
1 0 1
1 1 1
1 0 1
3 0 0
outputCopy
18
inputCopy
5 2
6 0 0
9 0 0
1 0 1
2 1 1
5 1 0
outputCopy
8
inputCopy
5 3
3 0 0
2 1 0
3 1 0
5 0 1
3 0 1
outputCopy
-1
#include<bits/stdc++.h>
using namespace std;
struct data{
int v,x,y;
}a[200010];
int b[200010],n,c[200010],k,xx,y,now,x,yy;
long long ans;
inline bool cmp(const data &x,const data &y){
return x.v<y.v;
}
int main(){
//freopen("e1.in","r",stdin);
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++) scanf("%d%d%d",&a[i].v,&a[i].x,&a[i].y);
x=y=xx=yy=ans=0,now=n;
sort(a+1,a+1+n,cmp);
for(int i=1;i<=n;i++){
if(a[i].x&&a[i].y){
if(xx<k&&yy<k) ++xx,++yy,ans+=a[i].v;
else if(yy>=k&&xx<k) ans-=c[y--],ans+=a[i].v,++xx;
else if(xx>=k&&yy<k) ans-=b[x--],ans+=a[i].v,++yy;
}else if(a[i].x&&xx<k) b[++x]=a[i].v,ans+=a[i].v,++xx;
else if(a[i].y&&yy<k) c[++y]=a[i].v,ans+=a[i].v,++yy;
if(xx>=k&&yy>=k){now=i;break;}
}
++now;
for(int i=now;i<=n;i++)
if(a[i].x&&a[i].y)
if(b[x]+c[y]>a[i].v) ans-=b[x--]+c[y--],ans+=a[i].v;
if(xx>=k&&yy>=k) printf("%lld\n",ans);
else printf("-1\n");
return 0;
}