Water and Jug Problem
<p>You are given two jugs with capacities <code>x</code> liters and <code>y</code> liters. You have an infinite water supply. Return whether the total amount of water in both jugs may reach <code>target</code> using the following operations:</p>

<ul>
	<li>Fill either jug completely with water.</li>
	<li>Completely empty either jug.</li>
	<li>Pour water from one jug into another until the receiving jug is full, or the transferring jug is empty.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> x = 3, y = 5, target = 4 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> true </span></p>

<p><strong>Explanation:</strong></p>

<p>Follow these steps to reach a total of 4 liters:</p>

<ol>
	<li>Fill the 5-liter jug (0, 5).</li>
	<li>Pour from the 5-liter jug into the 3-liter jug, leaving 2 liters (3, 2).</li>
	<li>Empty the 3-liter jug (0, 2).</li>
	<li>Transfer the 2 liters from the 5-liter jug to the 3-liter jug (2, 0).</li>
	<li>Fill the 5-liter jug again (2, 5).</li>
	<li>Pour from the 5-liter jug into the 3-liter jug until the 3-liter jug is full. This leaves 4 liters in the 5-liter jug (3, 4).</li>
	<li>Empty the 3-liter jug. Now, you have exactly 4 liters in the 5-liter jug (0, 4).</li>
</ol>

<p>Reference: The <a href="https://www.youtube.com/watch?v=BVtQNK_ZUJg&amp;ab_channel=notnek01" target="_blank">Die Hard</a> example.</p>
</div>

<p><strong class="example">Example 2: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> x = 2, y = 6, target = 5 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> false </span></p>
</div>

<p><strong class="example">Example 3: </strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> x = 1, y = 2, target = 3 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> true </span></p>

<p><strong>Explanation:</strong> Fill both jugs. The total amount of water in both jugs is equal to 3 now.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= x, y, target&nbsp;&lt;= 10<sup>3</sup></code></li>
</ul>
Depth-First Search; Breadth-First Search; Math
class Solution {
public:
    bool canMeasureWater(int x, int y, int z) {
        using pii = pair<int, int>;
        stack<pii> stk;
        stk.emplace(0, 0);
        auto hash_function = [](const pii& o) { return hash<int>()(o.first) ^ hash<int>()(o.second); };
        unordered_set<pii, decltype(hash_function)> vis(0, hash_function);
        while (stk.size()) {
            auto st = stk.top();
            stk.pop();
            if (vis.count(st)) {
                continue;
            }
            vis.emplace(st);
            auto [i, j] = st;
            if (i == z || j == z || i + j == z) {
                return true;
            }
            stk.emplace(x, j);
            stk.emplace(i, y);
            stk.emplace(0, j);
            stk.emplace(i, 0);
            int a = min(i, y - j);
            int b = min(j, x - i);
            stk.emplace(i - a, j + a);
            stk.emplace(i + b, j - b);
        }
        return false;
    }
};