You are given two arrays of length n: a1,a2,…,an and b1,b2,…,bn.
You can perform the following operation any number of times:
- Choose integer index i (1≤i≤n);
- Swap ai and bi.
What is the minimum possible sum |a1−a2|+|a2−a3|+⋯+|an−1−an| + |b1−b2|+|b2−b3|+⋯+|bn−1−bn| (in other words, ∑i=1n−1(|ai−ai+1|+|bi−bi+1|)) you can achieve after performing several (possibly, zero) operations?
Output
For each test case, print one integer — the minimum possible sum ∑i=1n−1(|ai−ai+1|+|bi−bi+1|).
Note
In the first test case, we can, for example, swap a3 with b3 and a4 with b4. We'll get arrays a=[3,3,3,3] and b=[10,10,10,10] with sum 3⋅|3−3|+3⋅|10−10|=0.
In the second test case, arrays already have minimum sum (described above) equal to |1−2|+⋯+|4−5|+|6−7|+⋯+|9−10| =4+4=8.
In the third test case, we can, for example, swap a5 and b5
.
The solution to the above question is
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define F first
#define S second
#define pb push_back
#define si set <int>
#define vi vector <int>
#define pii pair <int, int>
#define vpi vector <pii>
#define vpp vector <pair<int, pii>>
#define mii map <int, int>
#define mpi map <pii, int>
#define spi set <pii>
#define endl "\n"
#define sz(x) ((int) x.size())
#define all(p) p.begin(), p.end()
#define double long double
#define que_max priority_queue <int>
#define que_min priority_queue <int, vi, greater<int>>
#define bug(...) __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a) for(auto x : a) cout << x << " "; cout << endl
#define print1(a) for(auto x : a) cout << x.F << " " << x.S << endl
#define print2(a,x,y) for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl
inline int power(int a, int b)
{
int x = 1;
while (b)
{
if (b & 1) x *= a;
a *= a;
b >>= 1;
}
return x;
}
template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
const char* comma = strchr (names + 1, ',');
cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}
const int N = 200005;
void solve() {
int n;
cin>>n;
vi a(n,0),b(n,0);
for (int i = 0; i < n; i++)
{
cin>>a[i];
}
for (int i = 0; i < n; i++)
{
cin>>b[i];
}
int ans=0;
int noswap=0;
for (int i = 0; i < n-1; i++)
{
int curr=abs(a[i+1]-a[i])+abs(b[i+1]-b[i]);
int curr2=abs(a[i+1]-b[i])+abs(a[i]-b[i+1]);
if(curr2>curr){
swap(a[i],b[i]);
}
ans+=min(curr,curr2);
}
cout<<ans<<endl;
}
int32_t main()
{
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
clock_t z = clock();
int t = 1;
cin >> t;
while (t--) solve();
return 0;
}
Thank you for reading this.
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