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;
typedef pair<int, int> pii;

const int MN = 100010;

int N;
pii P[MN];
vector<pii> V;
int chk[MN];

int main() {
    scanf("%d", &N);

    for(int i = 0; i < N; i++) {
        scanf("%d %d", &P[i].first, &P[i].second);
        V.push_back(pii(P[i].first, 2 * i));
        V.push_back(pii(P[i].second, 2 * i + 1));
    }

    sort(V.begin(), V.end());

    ll sum = 0;
    for(int i = 0; i < N; i++) {
        sum += V[i].first;
    }
    if(N % 2) {

        bool bad1 = false;
        for(int i = 1; i < N; i++) {
            if(V[i].second % 2 != V[i - 1].second % 2) {
                bad1 = true;
                break;
            }
        }

        bool bad2 = true;
        for(int i = 0; i < N; i++) {
            if(chk[ V[i].second / 2 ]) {
                bad2 = false;
                break;
            }
            chk[ V[i].second / 2 ] = 1;
        }

        if(bad1 && bad2) {
            sum += V[N].first - V[N - 1].first;
        }
        printf("%lld", sum);
    }
    else printf("%lld", sum);
}