Evaluate Division

Evaluate Division

You are given an array of variable pairs equations and an array of real numbers values , where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i] . Each Ai or Bi is a string that represents a single variable.

You are also given some queries , where queries[j] = [Cj, Dj] represents the $j^{th}$ query where you must find the answer for Cj / Dj = ? .

Return the answers to all queries. If a single answer cannot be determined, return $-1.0$.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

Example 1:

Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]
Explanation: 
Given: a / b = 2.0, b / c = 3.0
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
return: [6.0, 0.5, -1.0, 1.0, -1.0 ]

Example 2:

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

Example 3:

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

 

解题思路

　　可以发现各个变量之间存在传递性，比如有$a = 2b,~ b = 3c$，那么一定会有$a = 6c$。因此想到用floyd求传递闭包。假设$f(a,b) = c$表示$a = c \cdot b$，那么如果存在一个$k$，有$f(a,k) = c_1,~ f(k,b) = c_2$，那么应该有$a = c_1 \cdot c_2 \cdot b$，即$f(a,b) = f(a,k) \times f(k,b)$。

　　AC代码如下：

class Solution {
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        unordered_map<string, int> mp;
        int n = 0;
        for (int i = 0; i < equations.size(); i++) {
            for (int j = 0; j < 2; j++) {
                if (!mp.count(equations[i][j])) mp[equations[i][j]] = ++n;
            }
        }

        vector<vector<double>> f(n + 1, vector<double>(n + 1, -1));
        for (int i = 1; i <= n; i++) {
            f[i][i] = 1;    // 一开始有x = 1 * x，因此f[x][x] = 1
        }

        for (int i = 0; i < equations.size(); i++) {
            int a = mp[equations[i][0]], b = mp[equations[i][1]];
            f[a][b] = values[i];        // a = val * b
            f[b][a] = 1 / values[i];    // b = 1/val * a
        }

        // floyd求传递闭包
        for (int k = 1; k <= n; k++) {
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    if (f[i][k] != -1 && f[k][j] != -1) f[i][j] = f[i][k] * f[k][j];    // 只有f[i][k]和f[k][j]存在才可以转移
                }
            }
        }

        vector<double> ans;
        for (auto &p : queries) {
            ans.push_back(f[mp[p[0]]][mp[p[1]]]);   // 如果p[0]或p[1]不存在，那么mp[p[0]]或mp[p[1]]得到的值是0，而f[0][x]或f[x][0]都为-1，(0 <= x <= n)
        }
        return ans;
    }
};

　　然后看了官方题解，发现还可以建图然后跑bfs。思路大概是把各个变量看作是结点，然后如果这两个变量之间存在关系，那么就连一条有向边，边的权值就是两个变量的比值。比如如果有$a = c \cdot b$，那么就在$a$到$b$连一条权值为$c$的边，在$b$到$a$连一条权值为$\frac{1}{c}$的边。

　　AC代码如下：

class Solution {
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        unordered_map<string, int> mp;
        int n = 0;
        for (int i = 0; i < equations.size(); i++) {
            for (int j = 0; j < 2; j++) {
                if (!mp.count(equations[i][j])) mp[equations[i][j]] = n++;
            }
        }

        // 建图
        vector<pair<int, double>> g[n];
        for (int i = 0; i < equations.size(); i++) {
            int a = mp[equations[i][0]], b = mp[equations[i][1]];
            g[a].push_back({b, values[i]});
            g[b].push_back({a, 1 / values[i]});
        }

        vector<double> ans;
        for (auto &p : queries) {
            if (!mp.count(p[0]) || !mp.count(p[1])) {   // 只要存在一个变量不在已知的关系中出现过，就返回-1
                ans.push_back(-1);
                continue;
            }

            queue<int> q({mp[p[0]]});   // 询问的是p[0]/p[1]，因此将p[0]加入队列
            vector<double> dist(n, -1); // dist[x]表示p[0]/x的值
            dist[mp[p[0]]] = 1; // p[0]/p[0] = 1
            while (!q.empty()) {
                int t = q.front();
                q.pop();

                for (auto &it : g[t]) {
                    if (dist[it.first] == -1) {
                        // dist[t]表示值p[0]/t，t到it.first存在一条权值为it.second的边，即有it.second = t/it.first
                        // 因此有dist[it.first] = p[0]/it.first = (p[0]/t) * (t/it.first) = dist[t] * it.second
                        dist[it.first] = dist[t] * it.second;
                        if (it.first == mp[p[1]]) break;
                        q.push(it.first);
                    }
                }
            }
            ans.push_back(dist[mp[p[1]]]);
        }

        return ans;
    }
};

　　甚至还可以用并查集，我都不知道怎么想到的。

　　应该是各个变量之间存在关系，因此可以放到一个集合，然后维护各个变量到根节点的距离，来表示集合中各个变量之间的关系。设$x$所在集合的代表元素（根节点）为$fa[x]$，定义$x$到根节点的距离$dist[x] = x / fa[x]$。

　　因此如果有询问求$a$和$b$的比值（关系），假设$a$和$b$在同一个集合中，根据定义有$dist[a] = a / fa[a],~ dist[b] = b / fa[b]$，其中$fa[a] = fa[b]$（因为$a$和$b$在同一个集合中），因此就有$a / b = dist[a] / dist[b]$。

　　已知条件$a / b = val$，现在要进行合并操作，首先应该让$fa[a]$（$a$所在集合的代表元素）指向$fa[b]$（$b$所在集合的代表元素），然后$dist[fa[a]]$的值应该是多少呢？首先根据定义应该有$dist[fa[a]] = fa[a] / fa[b]$，然后又有$dist[a] = a / fa[a]$（在调用find函数的时候已经路径压缩，$a$直接指向$fa[a]$，$b$同理），$dist[b] = b / fa[b]$，$a / b = dist[a] / dist[b]$，因此有

$$\begin{align*} dist[fa[a]] &= \frac{fa[a]}{fa[b]} \\ &= \frac{fa[a]}{a} \times \frac{a}{b} \times \frac{b}{fa[b]} \\ &= \frac{1}{dist[a]} \times val \times dist[b] \end{align*}$$

　　AC代码如下：

class Solution {
public:
    vector<int> fa;
    vector<double> dist;

    int find(int x) {
        if (fa[x] == x) return fa[x];
        int p = find(fa[x]);
        dist[x] *= dist[fa[x]];
        return fa[x] = p;
    }

    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        unordered_map<string, int> mp;
        int n = 0;
        for (int i = 0; i < equations.size(); i++) {
            for (int j = 0; j < 2; j++) {
                if (!mp.count(equations[i][j])) mp[equations[i][j]] = n++;
            }
        }

        for (int i = 0; i < n; i++) {
            fa.push_back(i);
            dist.push_back(1);
        }
        for (int i = 0; i < equations.size(); i++) {
            int a = mp[equations[i][0]], b = mp[equations[i][1]];
            int pa = find(a), pb = find(b);
            fa[pa] = pb;
            dist[pa] = values[i] * dist[b] / dist[a];
        }
        
        vector<double> ans;
        for (auto &p : queries) {
            if (!mp.count(p[0]) || !mp.count(p[1])) {
                ans.push_back(-1);
            }
            else {
                int a = mp[p[0]], b = mp[p[1]];
                if (find(a) == find(b)) ans.push_back(dist[a] / dist[b]);
                else ans.push_back(-1);
            }
        }

        return ans;
    }
};

 

参考资料　　

　　除法求值：https://leetcode.cn/problems/evaluate-division/solution/chu-fa-qiu-zhi-by-leetcode-solution-8nxb/