399. 除法求值(并查集)

399. 除法求值

给你一个变量对数组 equations 和一个实数值数组 values 作为已知条件，其中 equations[i] = [Ai, Bi] 和 values[i] 共同表示等式 Ai / Bi = values[i] 。每个 Ai 或 Bi 是一个表示单个变量的字符串。

另有一些以数组 queries 表示的问题，其中 queries[j] = [Cj, Dj] 表示第 j 个问题，请你根据已知条件找出 Cj / Dj = ? 的结果作为答案。

返回 所有问题的答案 。如果存在某个无法确定的答案，则用 -1.0 替代这个答案。如果问题中出现了给定的已知条件中没有出现的字符串，也需要用 -1.0 替代这个答案。

注意：输入总是有效的。你可以假设除法运算中不会出现除数为 0 的情况，且不存在任何矛盾的结果。

 

示例 1：

输入：equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
输出：[6.00000,0.50000,-1.00000,1.00000,-1.00000]
解释：
条件：a / b = 2.0, b / c = 3.0
问题：a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
结果：[6.0, 0.5, -1.0, 1.0, -1.0 ]

示例 2：

输入：equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
输出：[3.75000,0.40000,5.00000,0.20000]

示例 3：

输入：equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
输出：[0.50000,2.00000,-1.00000,-1.00000]

 

提示：

• 1 <= equations.length <= 20
• equations[i].length == 2
• 1 <= Ai.length, Bi.length <= 5
• values.length == equations.length
• 0.0 < values[i] <= 20.0
• 1 <= queries.length <= 20
• queries[i].length == 2
• 1 <= Cj.length, Dj.length <= 5
• Ai, Bi, Cj, Dj 由小写英文字母与数字组成

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <unordered_map>

using namespace std;

class Solution {
public:
    vector<int> parent; // 存储当前节点的根节点
    vector<double> weight; // 存储当前节点到根节点的权值
    void initUnionFind(int n) {
        parent.resize(n);
        weight.resize(n);
        for (int i = 0; i < n; i++) {
            parent[i] = i; // 初始每个集合只包含自己
            weight[i] = 1; // 自己与自己的商为1
        }
    }
    // 获取x与y的商
    double getTheQuotientOfTwoNumbers(int x, int y) {
        int xRoot = findRoot(x);
        int yRoot = findRoot(y);
        // 如果x与y在同一集合中,则获取x与y的商,否则不能获取到商
        if (xRoot == yRoot){
            return (weight[x] / weight[y]);
        } else {
            return -1.0;
        }
    }
    void unify(int x, int y, double value) {
        int xRoot = findRoot(x);
        int yRoot = findRoot(y);
        if (xRoot == yRoot) {
            return;
        }
        parent[xRoot] = yRoot;
        weight[xRoot] = (weight[y] * value) / weight[x]; // 合并两集合时,更新节点权值
        return;
    }
    int findRoot(int x) {
        if (x != parent[x]) {
            int next = parent[x];
            parent[x] = findRoot(parent[x]);
            weight[x] *= weight[next];
        }
        return parent[x];
    }
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        int n = equations.size();
        initUnionFind(2 * n);
        unordered_map<string, int> hashMap; // 存储字符串与数字的映射关系
        hashMap.reserve(2 * n);
        int id = 0;
        for (int i = 0; i < n; i++) {
            string str1 = equations[i][0];
            string str2 = equations[i][1];
            if (hashMap.count(str1) == 0) {
                hashMap[str1] = id;
                id++;
            }
            if (hashMap.count(str2) == 0) {
                hashMap[str2] = id;
                id++;
            }
            unify(hashMap[str1], hashMap[str2], values[i]);
        }
        cout << "hashMap:" << endl;
        for (auto &val : hashMap) {
            cout << "key:" << val.first << ", value:" << val.second << endl;
        }
        cout << "parent:" << endl;
        for (auto &val : parent) {
            cout << val << " ";
        }
        cout << endl;
        cout << "weight:" << endl;
        for (int i = 0; i < 2 * n; i++) {
            cout << "weight[" << i << "] = " << weight[i] << endl;
        }
        // 查询操作
        vector<double> ans;
        for (unsigned int i = 0; i < queries.size(); i++) {
            string str1 = queries[i][0];
            string str2 = queries[i][1];
            if (hashMap.count(str1) > 0 && hashMap.count(str2) > 0) {
                ans.push_back(getTheQuotientOfTwoNumbers(hashMap[str1], hashMap[str2]));
            } else {
                ans.push_back(-1.0);
            }
        }
        cout << "ans:" << endl;
        for (auto &val : ans) {
            cout << val << " ";
        }
        cout << endl;
        return ans;
    }
};

int main()
{
    /*
    *   [["x1","x2"],["x2","x3"],["x3","x4"],["x4","x5"]]
    *   [3.0,4.0,5.0,6.0]
    *   [["x1","x5"],["x5","x2"],["x2","x4"],["x2","x2"],["x2","x9"],["x9","x9"]]
    */
    Solution *test = new Solution();
    vector<vector<string>> equations = {
        {"x1", "x2"}, {"x2", "x3"}, {"x3", "x4"}, {"x4", "x5"}
    };
    vector<double> values = {3.0, 4.0, 5.0, 6.0};
    vector<vector<string>> queries = {
        {"x1", "x5"}, {"x5", "x2"}, {"x2", "x4"}, {"x2", "x2"}, {"x2", "x9"}, {"x9", "x9"}
    };
    vector<double> ans = test->calcEquation(equations, values, queries);

    std::cout << endl;
    return 0;
}