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;
typedef long long               ll;
const int mod = 1e9+7;
#ifndef M_PI
const double M_PI = acos(-1.0);
#endif
vector<int> ab[2];

int main(){
  int n;
  cin>>n;
  ab[0].resize(n);
  ab[1].resize(n);
  vector<bool> inact(n);
  priority_queue<pair<int, pair<bool, int > > > big[2];
  priority_queue<pair<int, pair<bool, int > >, vector< pair<int, pair<bool, int > > >, greater<pair<int, pair<bool, int > > > > small[2];
  for(int i=0; i<n; ++i){
    scanf("%d %d", &ab[0][i], &ab[1][i]);
    for(int j=0; j<2; ++j){
      big[j].emplace(ab[j][i], make_pair(j, i));
      small[j].emplace(ab[j][i], make_pair(j, i));
    }
  }
  ll res = 0;
  for(int i=0; i<n; ++i){
    for(int j=0; j<2; ++j){
      while(inact[small[j].top().second.second])
        small[j].pop();
    }
    bool j = small[0].top() > small[1].top();
    pair<int, pair<bool, int > > c_big, c_small;
    c_small = small[j].top();
    res += c_small.first;
    if(i != n-1){
      do{
        c_big = big[!j].top();
        big[!j].pop();
      } while(inact[c_big.second.second]);
      while(c_big.second.second == c_small.second.second){
        do{
          c_big = big[!j].top();
          big[!j].pop();
        } while(inact[c_big.second.second]);
      }
      inact[c_small.second.second] = true;
      inact[c_big.second.second] = true;
      ab[!j].push_back(ab[!j][c_small.second.second]);
      ab[j].push_back(ab[j][c_big.second.second]);
      inact.push_back(false);
      for(int k=0; k<2; ++k){
        big[k].emplace(ab[k].back(), make_pair(k, ab[k].size()-1));
        small[k].emplace(ab[k].back(), make_pair(k, ab[k].size()-1));
      }
    }
  }
  cout<<res<<'\n';
	return 0;
}