0399. 除法求值【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - 带权并查集

1. 📝 题目描述

• leetcode

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。

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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。

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注意：未在等式列表中出现的变量是未定义的，因此无法确定它们的答案。

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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 ]
注意：x 是未定义的 => -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]

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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

2. 🎯 s.1 - 带权并查集

c

// 带权并查集
char* names[1000];
double w[1000];
int par[1000];
int cnt;

int getId(char* s) {
    for (int i = 0; i < cnt; i++) if (strcmp(names[i], s) == 0) return i;
    names[cnt] = (char*)malloc(strlen(s) + 1);
    strcpy(names[cnt], s);
    par[cnt] = cnt; w[cnt] = 1.0;
    return cnt++;
}

int find(int x) {
    if (par[x] != x) {
        int root = find(par[x]);
        w[x] *= w[par[x]];
        par[x] = root;
    }
    return par[x];
}

void unite(int a, int b, double val) {
    int ra = find(a), rb = find(b);
    if (ra == rb) return;
    par[ra] = rb;
    w[ra] = val * w[b] / w[a];
}

double* calcEquation(char*** equations, int equationsSize, int* equationsColSize,
    double* values, int valuesSize, char*** queries, int queriesSize,
    int* queriesColSize, int* returnSize) {
    cnt = 0;
    for (int i = 0; i < equationsSize; i++) {
        int a = getId(equations[i][0]), b = getId(equations[i][1]);
        unite(a, b, values[i]);
    }
    double* res = (double*)malloc(sizeof(double) * queriesSize);
    *returnSize = queriesSize;
    for (int i = 0; i < queriesSize; i++) {
        int a = -1, b = -1;
        for (int j = 0; j < cnt; j++) {
            if (strcmp(names[j], queries[i][0]) == 0) a = j;
            if (strcmp(names[j], queries[i][1]) == 0) b = j;
        }
        if (a == -1 || b == -1 || find(a) != find(b)) res[i] = -1.0;
        else res[i] = w[a] / w[b];
    }
    for (int i = 0; i < cnt; i++) free(names[i]);
    return res;
}

js

/**
 * @param {string[][]} equations
 * @param {number[]} values
 * @param {string[][]} queries
 * @return {number[]}
 */
var calcEquation = function (equations, values, queries) {
  const parent = new Map()
  const weight = new Map() // weight[x] = x / root(x)

  const find = (x) => {
    if (!parent.has(x)) {
      parent.set(x, x)
      weight.set(x, 1.0)
    }
    if (parent.get(x) !== x) {
      const root = find(parent.get(x))
      weight.set(x, weight.get(x) * weight.get(parent.get(x)))
      parent.set(x, root)
    }
    return parent.get(x)
  }

  const union = (a, b, val) => {
    const ra = find(a),
      rb = find(b)
    if (ra === rb) return
    parent.set(ra, rb)
    weight.set(ra, (val * weight.get(b)) / weight.get(a))
  }

  for (let i = 0; i < equations.length; i++) {
    union(equations[i][0], equations[i][1], values[i])
  }

  return queries.map(([a, b]) => {
    if (!parent.has(a) || !parent.has(b)) return -1.0
    if (find(a) !== find(b)) return -1.0
    return weight.get(a) / weight.get(b)
  })
}

py

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        parent = {}
        weight = {}  # weight[x] = x / root(x)

        def find(x):
            if x not in parent:
                parent[x] = x
                weight[x] = 1.0
            if parent[x] != x:
                root = find(parent[x])
                weight[x] *= weight[parent[x]]
                parent[x] = root
            return parent[x]

        def union(a, b, val):
            ra, rb = find(a), find(b)
            if ra == rb:
                return
            parent[ra] = rb
            weight[ra] = val * weight[b] / weight[a]

        for (a, b), val in zip(equations, values):
            union(a, b, val)

        res = []
        for a, b in queries:
            if a not in parent or b not in parent or find(a) != find(b):
                res.append(-1.0)
            else:
                res.append(weight[a] / weight[b])
        return res

• 时间复杂度：$O((N+Q)\cdot \alpha (N))$，其中 $N$ 是等式数，$Q$ 是查询数
• 空间复杂度：$O(N)$

算法思路：

• 建立带权并查集，$weight[x]=x/root(x)$
• a / b = val 即合并 a、b，权值传递除法关系
• 查询时若同根则 $a/b=weight[a]/weight[b]$