Problem:
Score : 700 points

Problem Statement

We have a directed weighted graph with N vertices.
Each vertex has two integers written on it, and the integers written on Vertex i are A_i and B_i.

In this graph, there is an edge from Vertex x to Vertex y for all pairs 1 \leq x,y \leq N, and its weight is {\rm min}(A_x,B_y).

We will consider a directed cycle in this graph that visits every vertex exactly once.
Find the minimum total weight of the edges in such a cycle.

Constraints

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
A_2 B_2
:
A_N B_N

Output

Print the minimum total weight of the edges in such a cycle.

Sample Input 1

3
1 5
4 2
6 3

Sample Output 1

7

Consider the cycle 1→3→2→1. The weights of those edges are {\rm min}(A_1,B_3)=1, {\rm min}(A_3,B_2)=2 and {\rm min}(A_2,B_1)=4, for a total of 7.
As there is no cycle with a total weight of less than 7, the answer is 7.

Sample Input 2

4
1 5
2 6
3 7
4 8

Sample Output 2

10

Sample Input 3

6
19 92
64 64
78 48
57 33
73 6
95 73

Sample Output 3

227

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
template <class T> int size(const T &x) { return x.size(); }
#define rep(i,a,b) for (__typeof(a) i=(a); i<(b); ++i)
#define iter(it,c) for (__typeof((c).begin()) it = (c).begin(); it != (c).end(); ++it)
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef long long ll;
const int INF = ~(1<<31); // 2147483647

const double EPS = 1e-9;
const double pi = acos(-1);
typedef unsigned long long ull;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
template <class T> T smod(T a, T b) { return (a % b + b) % b; }

int main(){
    cin.sync_with_stdio(false);
    int n;
    cin >> n;
    vi a(n), b(n);
    vii ab(2*n);
    ll as = 0, bs = 0;
    rep(i,0,n){
        cin >> a[i] >> b[i];
        as += a[i];
        bs += b[i];
        ab[2*i] = ii(a[i], i);
        ab[2*i+1] = ii(b[i], i);
    }
    sort(ab.begin(), ab.end());
    ll sm = 0;
    ll ans = min(as, bs);
    vector<int> used(n, 0);
    rep(i,0,n-1){
        sm += ab[i].first;
        used[ab[i].second]++;
    }
    rep(i,0,n){
        if(used[i] == 2){
            ans = min(ans, sm+ab[n-1].first);
        }
        else if(used[i] == 1){
            ans = min(ans, sm+max(a[i], b[i]));
        }
        else{
            ans = min(ans, sm+a[i]+b[i]-ab[n-2].first);
        }
    }
    cout << ans << endl;
    return 0;
}