problem
D. Phoenix and Socks
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
To satisfy his love of matching socks, Phoenix has brought his n socks (n is even) to the sock store. Each of his socks has a color ci and is either a left sock or right sock.
Phoenix can pay one dollar to the sock store to either:
recolor a sock to any color c′ (1≤c′≤n)
turn a left sock into a right sock
turn a right sock into a left sock
The sock store may perform each of these changes any number of times. Note that the color of a left sock doesn’t change when it turns into a right sock, and vice versa.
A matching pair of socks is a left and right sock with the same color. What is the minimum cost for Phoenix to make n/2 matching pairs? Each sock must be included in exactly one matching pair.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤t≤1000) — the number of test cases.
The first line of each test case contains three integers n, l, and r (2≤n≤2⋅105; n is even; 0≤l,r≤n; l+r=n) — the total number of socks, and the number of left and right socks, respectively.
The next line contains n integers ci (1≤ci≤n) — the colors of the socks. The first l socks are left socks, while the next r socks are right socks.
It is guaranteed that the sum of n across all the test cases will not exceed 2⋅105.
Output
For each test case, print one integer — the minimum cost for Phoenix to make n/2 matching pairs. Each sock must be included in exactly one matching pair.
Example
inputCopy
4
6 3 3
1 2 3 2 2 2
6 2 4
1 1 2 2 2 2
6 5 1
6 5 4 3 2 1
4 0 4
4 4 4 3
outputCopy
2
3
5
3
Note
In the first test case, Phoenix can pay 2 dollars to:
recolor sock 1 to color 2
recolor sock 3 to color 2
There are now 3 matching pairs. For example, pairs (1,4), (2,5), and (3,6) are matching.
In the second test case, Phoenix can pay 3 dollars to:
turn sock 6 from a right sock to a left sock
recolor sock 3 to color 1
recolor sock 4 to color 1
There are now 3 matching pairs. For example, pairs (1,3), (2,4), and (5,6) are matching.
solution
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+10;
int lc[maxn], rc[maxn];
int main(){
int T; cin>>T;
while(T--){
memset(lc,0,sizeof(lc));
memset(rc,0,sizeof(rc));
int n, l, r; cin>>n>>l>>r;
for(int i = 1; i <= n; i++){
int c; cin>>c;
if(i<=l)lc[c]++;
else rc[c]++;
}
int ld = l-n/2, rd = r-n/2;
int lr = 0, rr = 0;
int ans = 0;
for(int i = 1; i <= n; i++){
if(lc[i]==rc[i])continue;
if(lc[i]>rc[i]){
lc[i] -= rc[i];//0配对
while(lc[i]>=2 && ld>0){//1变形状
lc[i] -= 2;
ld--;
ans++;
}
//if(ld<=0){//1变颜色
lr += lc[i];
//}
}else{
rc[i] -= lc[i];
while(rc[i]>=2 && rd>0){
rc[i] -= 2;
rd--;
ans++;
}
//if(rd<=0){
rr += rc[i];
//}
}
}
//cout<<ans<<":"<<lr<<","<<rr<<"\n";
//cout<<ld<<" "<<rd<<"\n";
if(ld>0 || rd>0){
if(ld>0){
ans += 2*ld;
//rr -= ld;
//ld = 0;
lr -= 2*ld;
//ans += lr;
}
if(rd>0){
ans += 2*rd;
//lr -= rd;
//rd = 0;
rr -= 2*rd;
//ans += rr;
}
}
if(lr!=0 && rr!=0){
ans += lr;
}
cout<<ans<<"\n";
}
return 0;
}