365. Water and Jug Problem(Leetcode每日一题-2020.03.21)

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

• Fill any of the jugs completely with water.
• Empty any of the jugs.
• Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example1

Input: x = 3, y = 5, z = 4
Output: True

Example2

Input: x = 2, y = 6, z = 5
Output: False

Solution

stack+DFS+unordered_set记录已经遍历过的状态以避免无限递归

using PII = pair<int, int>;

struct pair_hash
{
	template <class T1, class T2>
	std::size_t operator () (std::pair<T1, T2> const &pair) const
	{
		std::size_t h1 = std::hash<T1>()(pair.first);
		std::size_t h2 = std::hash<T2>()(pair.second);

		return h1 ^ h2;
	}
};

class Solution {
public:
    bool canMeasureWater(int x, int y, int z) {
        stack<PII> stk;
        stk.emplace(0, 0);
       
        unordered_set<PII, pair_hash> seen;
        while (!stk.empty()) {
            if (seen.count(stk.top())) {
                stk.pop();
                continue;
            }
            seen.emplace(stk.top());
            
            auto [remain_x, remain_y] = stk.top();
            stk.pop();
            if (remain_x == z || remain_y == z || remain_x + remain_y == z) {
                return true;
            }
            // 把 X 壶灌满。
            stk.emplace(x, remain_y);
            // 把 Y 壶灌满。
            stk.emplace(remain_x, y);
            // 把 X 壶倒空。
            stk.emplace(0, remain_y);
            // 把 Y 壶倒空。
            stk.emplace(remain_x, 0);
            // 把 X 壶的水灌进 Y 壶，直至灌满或倒空。
            stk.emplace(remain_x - min(remain_x, y - remain_y), remain_y + min(remain_x, y - remain_y));
            // 把 Y 壶的水灌进 X 壶，直至灌满或倒空。
            stk.emplace(remain_x + min(remain_y, x - remain_x), remain_y - min(remain_y, x - remain_x));
        }
        return false;
    }
};