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;
int32_t main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n; cin >> n;
    vector< pair<int, int> > vec(2*n);
    for (int i = 0; i < n; ++i) {
		vec[i].second = i, vec[i+n].second = i+n;
		cin >> vec[i].first >> vec[i+n].first;
    }
    sort(vec.begin(), vec.end());

    bool hasA = 0, hasB = 0, hasD = 0;
    vector<bool> used(n, 0);
    int64_t ans = 0;
    for (int i = 0; i < n; ++i) {
        ans += vec[i].first;
        if (vec[i].second < n) hasA = 1; else hasB = 1;
        if (used[vec[i].second % n]) hasD = 1; else used[vec[i].second % n] = 1;
    }
    if (hasA && hasB && (!hasD)) {
        ans -= vec[n-1].first - vec[n].first;
    }
    cout << ans;
}