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>
#define sz(x) ((int)x.size())
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef long double ld;
int n, A[100010], B[100010], P[100010][2];
ll s[200010];
vector<pair<int,int> > v;
int main() {
	memset(P, -1, sizeof(P));
	scanf("%d", &n);
	ll r1 = 0, r2 = 0;
	for(int i=1; i<=n; i++) {
		scanf("%d%d", A+i, B+i);
		r1 += A[i]; r2 += B[i];
		v.push_back({A[i], i});
		v.push_back({B[i], i});
	}
	sort(all(v));
	for(int i=0; i<sz(v); i++) {
		s[i] += v[i].first;
		s[i+1] += s[i];
		if(P[v[i].second][0] == -1) P[v[i].second][0] = i;
		else P[v[i].second][1] = i;
	}
	ll r = min(r1, r2);
	for(int i=1; i<=n; i++) {
		ll val = 0;
		if(n == 2);
		else if(n-3 < P[i][0]) val = s[n-3];
		else if(n-3 >= P[i][1]) val = s[n-1] - A[i] - B[i];
		else if(n-2 == P[i][1]) val = s[n-1] - A[i] - B[i];
		else val = s[n-2] - v[P[i][0]].first;
		r = min(r, val + A[i] + B[i]);
	}
	printf("%lld", r);
	return 0;
}