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 max_n = 100111, inf = 1000111222;

int n, a[max_n], b[max_n];
multiset<pair<int, int>> xy, yx;

void add(int x, int y) {
    //cout << "+ " << x << " " << y << endl;
    xy.insert({x, y});
    yx.insert({y, x});
}

void del(int x, int y) {
    //cout << "- " << x << " " << y << endl;
    xy.erase(xy.find({x, y}));
    yx.erase(yx.find({y, x}));
}

vector<int> all;

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
        scanf("%d%d", &a[i], &b[i]);
        add(a[i], b[i]);
        all.push_back(a[i]);
        all.push_back(b[i]);
    }
    sort(all.begin(), all.end());
    long long qqq = 0;
    for (int i = 0; i < n; ++i) {
        qqq += all[i];
    }
    long long ans = 0;
    while (xy.size() > 1) {
        pair<int, int> p1 = *xy.begin();
        pair<int, int> p2 = *yx.begin();
        int x, y, tp;
        if (p1.first < p2.first) {
            x = p1.first;
            y = p1.second;
            tp = 0;
        } else {
            x = p2.second;
            y = p2.first;
            tp = 1;
        }
        ans += min(x, y);
        //cout << x << " " << y << endl;
        del(x, y);
        if (tp == 0) {
            pair<int, int> p = *yx.rbegin();
            del(p.second, p.first);
            add(p.second, y);
        } else {
            pair<int, int> p = *xy.rbegin();
            del(p.first, p.second);
            add(x, p.second);
        }
    }
    ans += min(xy.begin()->first, xy.begin()->second);
    if (ans != qqq) {
        qqq -= all[n - 1];
        qqq += all[n];
        ans = qqq;
    }
    cout << ans << "\n";
    return 0;
}