目录 题目链接 注意点 解法 小结 题目链接 Satisfiability of Equality Equations - LeetCode 注意点 必须要初始化pre 解法 解法一:典型的并查集算法应用.先遍历所有等式,将等号两边的字母加入同一分类,每类中的字母都是相等的.然后遍历不等式,如果不等号两边的字母属于同一类则返回false.时间复杂度O(nm) class Solution { public: map<char,char> pre; char Find(char x) { cha…
Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily dif…
原题链接在这里:https://leetcode.com/problems/satisfiability-of-equality-equations/ 题目: Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b"…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 并查集 日期 题目地址:https://leetcode.com/problems/satisfiability-of-equality-equations/ 题目描述 Given an array equations of strings that represent relationships between variables, eac…
题目如下: Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and bare lowercase letters (not necessaril…
Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily diff…
Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily dif…
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Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0. Example:Given a / b =…
我是链接 看到这道题,2个点和一个权值,然后想到图,但是leetcode就是这样,没给数据范围,感觉写起来很费劲,然后就开始用图来做,添加边的时候,注意正向边和反向变,然后查询的时候,先判断2个点是否都出现,然后判断是不是自己到自己,最后用dfs或者bfs到图里面去搜索,第一次交的时候,忘记用vis标记访问过的点,导致tle,加上之后,就ac了,套路一般吧,不知道还有什么trick. double dfs(vector<vector<pair<int, double> > &…