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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
#define int long long
typedef pair<int, int> P;
typedef pair<int, P> Pi;
 
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);
 
  int N;
  cin >> N;
 
  vector<int> A(N), B(N);
  vector<Pi> C; 
  for ( int i = 0; i < N; i++ ) {
    cin >> A[i] >> B[i];
    C.emplace_back(Pi{A[i], P{0, i}});
    C.emplace_back(Pi{B[i], P{1, i}});
  }
 
  sort(C.begin(), C.end());
  int cnt = 0, sum = 0;
  int k = 0;
  int c1 = 0, c2 = 0;  
  bool used[100001] = {0};
  bool f = false;  
  for ( int i = 0; i < (int)C.size(); i++ ) {
    int c = C[i].first;
    int d = C[i].second.first;
    int id = C[i].second.second;
    if ( d == 0 ) {
      if ( c1 != N-1 ) {
	if ( cnt == N-1 && !f ) continue;	
      }
    } else {
      if ( c2 != N-1 ) {
	if ( cnt == N-1 && !f ) continue;	
      }
    }
    // cout << c << endl;
    if ( !used[id] ) k++;
    else f = true;
    used[id] = true;
    cnt++;
    sum += c;
    if ( d == 0 ) c1++;
    if ( d == 1 ) c2++;
    if ( cnt >= N ) break;    
  }
 
  cout << sum << endl;
}