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;
typedef long long ll;
const ll MOD = 1e9 + 7;
const double EPS = 1e-10;
typedef pair<ll, int> P;
ll A[100000];
ll B[100000];
int cnt[100000];
int main() {
	int N;
	cin >> N;
	vector<P> vec;
	ll ans = 1LL << 60;
	ll sumA = 0, sumB = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i] >> B[i];
		sumA += A[i];
		sumB += B[i];
		vec.emplace_back(A[i], i);
		vec.emplace_back(B[i], i);
	}
	ans = min(ans, sumA);
	ans = min(ans, sumB);
	sort(vec.begin(), vec.end());
	for (int i = 0; i < N; i++) cnt[vec[i].second]++;
	bool exist = 0;
	for (int i = 0; i < N; i++) {
		if (cnt[i] == 2) {
			exist = 1;
			break;
		}
	}
	ll sum = 0;
	for (int i = 0; i < N; i++) {
		sum += vec[i].first;
	}
	if (exist) {
		ans = min(ans, sum);
	}
	else {
		for (int i = 0; i < N; i++) {
			ll t = sum;
			if (A[i] <= B[i]) {
				t -= A[i];
				if (vec[N].second != i) t += vec[N].first;
				else t += vec[N + 1].first;
			}
			else {
				t -= B[i];
				if (vec[N].second != i) t += vec[N].first;
				else t += vec[N + 1].first;
			}
			ans = min(ans, t);
		}
	}
	cout << ans << endl;
}