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;

const int N = 100000;

int n;
int a[N + 5], b[N + 5];
vector<pair<int, int> > val;
long long ans = LLONG_MAX, sa, sb;

int main() {

    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d%d", a+i, b+i), sa += a[i], sb += b[i];
        val.emplace_back(a[i], i);
        val.emplace_back(b[i], i);
    }
    ans = min(sa, sb);

    std::sort(val.begin(), val.end());

    long long s0 = 0, s1 = 0, s2 = 0;
    for(int i = 0; i < n-2; ++i) {
        s0 += val[i].first; 
    }
    s1 = s0 + val[n-2].first;
    s2 = s1 + val[n-1].first;

    static int vis[N + 5];
    memset(vis, -1, sizeof vis);

    for(int i = 0; i < 2*n; ++i) {
        if(vis[val[i].second] >= 0) {
            int c = (i < n) + (vis[val[i].second] < n);
            if(c == 2) ans = min(ans, s2);
            if(c == 1) ans = min(ans, val[i].first + s1);
            if(c == 0) ans = min(ans, val[i].first + val[vis[val[i].second]].first + s0);
        } else vis[val[i].second] = i;
    }
    printf("%lld\n", ans);

    return 0;
}