zoukankan      html  css  js  c++  java
  • LC 802. Find Eventual Safe States

    In a directed graph, we start at some node and every turn, walk along a directed edge of the graph.  If we reach a node that is terminal (that is, it has no outgoing directed edges), we stop.

    Now, say our starting node is eventually safe if and only if we must eventually walk to a terminal node.  More specifically, there exists a natural number K so that for any choice of where to walk, we must have stopped at a terminal node in less than K steps.

    Which nodes are eventually safe?  Return them as an array in sorted order.

    The directed graph has N nodes with labels 0, 1, ..., N-1, where N is the length of graph.  The graph is given in the following form: graph[i] is a list of labels jsuch that (i, j) is a directed edge of the graph.

    Example:
    Input: graph = [[1,2],[2,3],[5],[0],[5],[],[]]
    Output: [2,4,5,6]
    Here is a diagram of the above graph.
    

    Runtime: 268 ms, faster than 12.50% of C++ online submissions for Find Eventual Safe States.

    slow

    class Solution {
    public:
      vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        vector<int> indegree(graph.size(),0);
        vector<int> outdegree(graph.size(), 0);
        unordered_map<int,vector<int>> parent;
        for(int i=0; i<graph.size(); i++){
          for(int j=0; j<graph[i].size(); j++){
            indegree[graph[i][j]]++;
            outdegree[i]++;
            parent[graph[i][j]].push_back(i);
          }
        }
        queue<int> q;
        unordered_map<int,bool> used;
        for(int i=0; i<graph.size(); i++) used[i] = false;
        while(true) {
          for(int i=0; i<outdegree.size(); i++) {
            if(outdegree[i] == 0 && !used[i]) {
              q.push(i);
            }
          }
          if(q.empty()) break;
          while(!q.empty()) {
            int tmp = q.front(); q.pop();
            used[tmp] = true;
            for(int x : parent[tmp]) {
              outdegree[x]--;
            }
          }
        }
        vector<int> ret;
        for(int i=0; i<outdegree.size(); i++){
          if(outdegree[i] == 0) ret.push_back(i);
        }
        return ret;
      }
    };

    Runtime: 140 ms, faster than 100.00% of C++ online submissions for Find Eventual Safe States.

    class Solution {
    
    public:
      vector<int> eventualSafeNodes(vector<vector<int>>& graph) {
        vector<int> color(graph.size(),0);
        vector<int> ret;
        for(int i=0; i<graph.size(); i++){
          if(dfs(graph, i, color)) ret.push_back(i);
        }
        return ret;
      }
    
      bool dfs(vector<vector<int>>& graph, int s, vector<int>& color) {
        if(color[s] > 0) return color[s] == 2;
        color[s] = 1;
        for(int& x : graph[s]) {
          if(!dfs(graph, x, color)) return false;
        }
        color[s] = 2;
        return true;
      }
    };
  • 相关阅读:
    Python 函数 I
    jmeter-将上一个接口的返回值作为下一个接口的参数
    Python 文件的操作
    Python 基础数据类型 VI
    Pyhton 基础数据类型 V (补充)
    Python 基础数据类型 IV (集合)
    Python 基础数据类型 III (字典)
    难道是你?
    是你啦
    checkWeb
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10285718.html
Copyright © 2011-2022 走看看