E. The Contest
A team of three programmers is going to play a contest. The contest consists of 𝑛 problems, numbered from 1 to 𝑛. Each problem is printed on a separate sheet of paper. The participants have decided to divide the problem statements into three parts: the first programmer took some prefix of the statements (some number of first paper sheets), the third contestant took some suffix of the statements (some number of last paper sheets), and the second contestant took all remaining problems. But something went wrong — the statements were printed in the wrong order, so the contestants have received the problems in some random order.
The first contestant has received problems 𝑎1,1,𝑎1,2,…,𝑎1,𝑘1. The second one has received problems 𝑎2,1,𝑎2,2,…,𝑎2,𝑘2. The third one has received all remaining problems (𝑎3,1,𝑎3,2,…,𝑎3,𝑘3).
The contestants don't want to play the contest before they redistribute the statements. They want to redistribute them so that the first contestant receives some prefix of the problemset, the third contestant receives some suffix of the problemset, and the second contestant receives all the remaining problems.
During one move, some contestant may give one of their problems to other contestant. What is the minimum number of moves required to redistribute the problems?
It is possible that after redistribution some participant (or even two of them) will not have any problems.
Input
The first line contains three integers 𝑘1,𝑘2 and 𝑘3 (1≤𝑘1,𝑘2,𝑘3≤2⋅105,𝑘1+𝑘2+𝑘3≤2⋅105) — the number of problems initially taken by the first, the second and the third participant, respectively.
The second line contains 𝑘1 integers 𝑎1,1,𝑎1,2,…,𝑎1,𝑘1 — the problems initially taken by the first participant.
The third line contains 𝑘2 integers 𝑎2,1,𝑎2,2,…,𝑎2,𝑘2 — the problems initially taken by the second participant.
The fourth line contains 𝑘3 integers 𝑎3,1,𝑎3,2,…,𝑎3,𝑘3 — the problems initially taken by the third participant.
It is guaranteed that no problem has been taken by two (or three) participants, and each integer 𝑎𝑖,𝑗 meets the condition 1≤𝑎𝑖,𝑗≤𝑛, where 𝑛=𝑘1+𝑘2+𝑘3.
Output
Print one integer — the minimum number of moves required to redistribute the problems so that the first participant gets the prefix of the problemset, the third participant gets the suffix of the problemset, and the second participant gets all of the remaining problems.
Examples
input
2 1 2
3 1
4
2 5
output
1
3 2 1
5 4
6
2 1 3
5 6
1 2 3
3
1 5 1
5 1 2 4 7
2
Note
In the first example the third contestant should give the problem 2 to the first contestant, so the first contestant has 3 first problems, the third contestant has 1 last problem, and the second contestant has 1 remaining problem.
In the second example the distribution of problems is already valid: the first contestant has 3 first problems, the third contestant has 1 last problem, and the second contestant has 2 remaining problems.
The best course of action in the third example is to give all problems to the third contestant.
The best course of action in the fourth example is to give all problems to the second contestant.
題意
現在有三個隊列，每個隊列都有一堆的數。現在你要使得a隊列裡面的數為排列的字首，c隊列的數是排列的字尾，然後b隊列是其他的數。每次操作你可以讓一個數到另外的隊列裡面去，問你最少操作多少次，能夠符合需求。
題解
我們考慮排完序之後拼在一起，那麼這個序列的lis，就是最後能夠保持不變的。
那麼nlogn跑一個lis就行。
代碼
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
int k1,k2,k3,n,a[maxn],dp[maxn],b[maxn],len;
void lis(){
len=1;b[1]=a[0];
for(int i=1;i<n;i++){
b[len+1]=n+1;
int l=0,r=len+1,ans=0;
while(l<=r){
int mid=(l+r)/2;
if(a[i]<b[mid]){
ans=mid;
r=mid-1;
}else{
l=mid+1;
}
}
b[ans]=a[i];
if(ans>len)len++;
}
cout<<n-len<<endl;
}
int main(){
scanf("%d%d%d",&k1,&k2,&k3);
n=k1+k2+k3;
for(int i=0;i<k1+k2+k3;i++){
scanf("%d",&a[i]);
}
sort(a,a+k1);
sort(a+k1,a+k1+k2);
sort(a+k1+k2,a+k1+k2+k3);
lis();
}