Old legos

input/code.cpp: In function 'int main(int, char**)':
input/code.cpp:122:23: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<std::pair<int, int> >::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  122 |     for (int i = 0; i < UM.size(); ++i) {
      |                     ~~^~~~~~~~~~~

#include <algorithm>
#include <iostream>
#include <limits>
#include <utility>
#include <vector>

/**
Uolevi and Maija are visiting their parents. While the blueberry pie is cooking
in the oven, they go explore the attic. They find their old legos and decide to
build castles from them. There are n legos in a box numbered 1,2,\dots,n, the
i-th of which has u_i nostalgic value to Uolevi and m_i to Maija. The siblings
then take turns in picking legos from the box and placing them to their castles.
Naturally, both try to maximize the value difference between their own castle
and their opponents castle. Formally, the legos will be divided into two sets U
and M at the end of the process. Uolevi's castle has value x = \sum_{i \in
U}{u_i} and Maija's castle has value y = \sum_{i \in M}{m_i}. Maija tries to
minimize x-y while Uolevi tries to maximize it. Calculate x-y if Uolevi gets to
pick the first lego and both pick the legos optimally.

Input
First line contain a single integer n.
The 𝑖-th of the next n lines contains two integers u_i and m_𝑖.

Output
Output a single integer x-y.
Constraints

1 \leq n \leq 10^5
1 \leq u_i, m_i \leq 10^9

Example 1
Input:
5
3 1
9 3
6 3
2 10
1 7

Output:
1

Explanation:
The sequence of picks would be:
Uolevi picks lego 2
Maija picks lego 4
Uolevi picks lego 3
Maija picks lego 5
Uolevi picks lego 1
Example 2
Input:
2
10 2
1 1000

Output:
-1
 */

using namespace std;

const int INF = numeric_limits<int>::max();

template <typename T>
std::ostream &operator<<(std::ostream &os, const vector<T> &obj) {
    bool first = true;
    for (const auto &o : obj) {
        if (first) {
            cout << o;
            first = false;
        } else {
            cout << ' ' << o;
        }
    }
    cout << endl;
    return os;
}

std::ostream &operator<<(std::ostream &os, const vector<pair<int, int>> &obj) {
    for (const auto &o : obj) {
        cout << o.first << ' ' << o.second << endl;
    }
    cout << endl;
    return os;
}

// vector<int> U;
// vector<int> M;
// vector<int> x_y;

int main(int argc, char **argv) {
    int n;
    cin >> n;

    // U = vector<int>(n);
    // M = vector<int>(n);
    // x_y = vector<int>(n);

    auto UM = vector<pair<int, int>>(n);

    for (int i = 0; i < n; ++i) {
        cin >> UM[i].first;
        cin >> UM[i].second;
        // cin >> U[i];
        // cin >> M[i];
        // x_y[i] = U[i] - M[i];
    }

    // sort(U.begin(), U.end());
    // sort(M.begin(), M.end());
    // sort(x_y.begin(), x_y.end());
    sort(UM.begin(), UM.end(), [](pair<int, int> a, pair<int, int> b) {
        return abs(a.first - a.second) > abs(b.first - b.second);
    });

    cout << endl;
    cout << "UM: " << endl << UM << endl;
    // cout << x_y << endl;
    // int m = 0, u = x_y.size() - 1;
    char turn{'U'};
    int x_y = 0;
    for (int i = 0; i < UM.size(); ++i) {

        if (turn == 'U') {
            x_y += UM[i].first;
            turn = 'M';
        } else {
            x_y -= UM[i].second;
            turn = 'U';
        }
    }
    // while () {
    //     switch (turn) {
    //     case 'U':
    //         choice = x_y[u--];
    //         d += choice;
    //         break;
    //     case 'M':
    //         choice = x_y[m++];
    //         d -= choice;
    //         break;
    //     }
    //     cout << choice << endl;
    // }
    cout << x_y << endl;
}

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