图 floyd

399. Evaluate Division

Medium

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 jth 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] 

Constraints:

• 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 consist of lower case English letters and digits.

解法一：bfs

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        Map<String,List<Pair>> graph = new HashMap();
        //O(E)
        for(int i=0;i<values.length;i++){
            String a = equations.get(i).get(0);
            String b = equations.get(i).get(1);
            List<Pair> list = graph.getOrDefault(a,new ArrayList());
            list.add(new Pair(b,values[i]));
            graph.put(a, list);
            list = graph.getOrDefault(b,new ArrayList());
            list.add(new Pair(a,1/values[i]));
            graph.put(b, list);
        }
        double[] result = new double[queries.size()];
        //O(VN)
        for(int i=0;i<queries.size();i++){
            result[i] = bfs(graph, queries.get(i).get(0), queries.get(i).get(1) );
        }
        
        //summaray: O(E+EN) ??
        return result;
    }
    //O(V)
    private double bfs(Map<String,List<Pair>> graph, String start, String end){
        if(!graph.containsKey(start) || !graph.containsKey(end)) return -1.0;
        Queue<Pair> queue = new LinkedList();
        Set<String> visited = new HashSet();
        queue.offer(new Pair(start,1));
        visited.add(start);
        while(!queue.isEmpty()){
            Pair curr = queue.poll();
            if(curr.key.equals(end)) return curr.val;
            for(Pair other:graph.getOrDefault(curr.key,List.of())){
                if(visited.contains(other.key)) continue;
                visited.add(other.key);
                queue.offer(new Pair(other.key, curr.val*other.val));
            }
        }
        return -1;
    }
}
class Pair{
    String key;
    double val;
    Pair(String key, double val){
        this.key = key;
        this.val = val;
    }
}

 

解法二：floyd

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        Map<String,Map<String,Double>> map = new HashMap();
        
        for(int i=0;i<values.length;i++){
            String first = equations.get(i).get(0),second = equations.get(i).get(1);
            map.put(first,map.getOrDefault(first,new HashMap()));
            map.put(second,map.getOrDefault(second,new HashMap()));
            Map<String,Double> firstMap = map.get(first);
            Map<String,Double> secondMap = map.get(second);
            firstMap.put(first,1d);//便于handle a/a的运算
            secondMap.put(second,1d);
            firstMap.put(second,values[i]);
            secondMap.put(first,1/values[i]);
        }
        for(String key:map.keySet()){
            Map<String,Double> currmap = map.get(key);
            for(String key1:currmap.keySet()){
                for(String key2:currmap.keySet()){
                    if(key.equals(key1) || key.equals(key2) || key1.equals(key2)) continue;
                    double value12 = map.get(key1).get(key)/map.get(key2).get(key);
                    map.get(key1).put(key2,value12);
                    map.get(key2).put(key1,1/value12);
                }
            }
        }
        double[] results = new double[queries.size()];
        for(int i=0;i<queries.size();i++){
            String first = queries.get(i).get(0),second = queries.get(i).get(1);
            if(map.get(first)!=null && map.get(first).get(second)!=null) //处理 a/b  但a,b 不存在关系的情况
                results[i]=map.get(first).get(second);
            else
                results[i]=-1;
        }
        return results;
    }
}

 

1462. Course Schedule IV

Medium

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There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course ai first if you want to take course bi.

• For example, the pair [0, 1] indicates that you have to take course 0 before you can take course 1.

Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c.

You are also given an array queries where queries[j] = [uj, vj]. For the jth query, you should answer whether course uj is a prerequisite of course vj or not.

Return a boolean array answer, where answer[j] is the answer to the jth query.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: The pair [1, 0] indicates that you have to take course 1 before you can take course 0.
Course 0 is not a prerequisite of course 1, but the opposite is true.

Example 2:

Input: numCourses = 2, prerequisites = [], queries = [[1,0],[0,1]]
Output: [false,false]
Explanation: There are no prerequisites, and each course is independent.

Example 3:

Input: numCourses = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
Output: [true,true]

Constraints:

• 2 <= numCourses <= 100
• 0 <= prerequisites.length <= (numCourses * (numCourses - 1) / 2)
• prerequisites[i].length == 2
• 0 <= ai, bi <= n - 1
• ai != bi
• All the pairs [ai, bi] are unique.
• The prerequisites graph has no cycles.
• 1 <= queries.length <= 104
• 0 <= ui, vi <= n - 1
• ui != vi

class Solution {
    public List<Boolean> checkIfPrerequisite(int numCourses, int[][] pre, int[][] queries) {
        //1.建图
        Map<Integer,List<Integer>> graph = new HashMap();
        int[] indegree = new int[numCourses];
        for(int[] pair:pre){
            List<Integer> list = graph.getOrDefault(pair[0],new ArrayList());
            graph.put(pair[0],list);
            list.add(pair[1]);
            indegree[pair[1]]++;
        }
        //2.将入度为0的点加入队列
        List<Integer> parents = new ArrayList();
        boolean[][] mem = new boolean[numCourses][numCourses];
        Queue<Integer> queue = new LinkedList();
        for(int i=0;i<numCourses;i++){
            if(indegree[i]==0) queue.offer(i);
        }
        //3.bfs 拓扑遍历
        while(!queue.isEmpty()){
            int curr = queue.poll();
            //加入已遍历列表
            parents.add(curr);
            for(int neighbor:graph.getOrDefault(curr,Arrays.asList())){
                indegree[neighbor]--;
                if(indegree[neighbor]==0) queue.offer(neighbor);
                //设置当前点与其neighbor关系
                mem[curr][neighbor]=true;
                //判断当前点与已遍历节点的关系，传递给已遍历点到neighbor
                for(int parent:parents){
                    if(mem[parent][curr]) mem[parent][neighbor]=true;
                }
            }
        }
        //4.query 得到结果集
        List<Boolean> result = new ArrayList();
        for(int[] pair:queries)  result.add(mem[pair[0]][pair[1]]);
        
        return result;
    }
}