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:
    Eclipse中properties文件中文显示编码、乱码问题
    Eclipse中安装yml插件( YEdit )
    Java中如何返回Json数组
    ASIFormDataRequest 登录
    Safari里使用JsonView
    beginUpdates和endUpdates
    svn log 不显示日志的问题
    svn代码回滚命令
    Tomcat: localhost:8080 提示404
    android定时三种方式
  • 原文地址:https://www.cnblogs.com/ethanhong/p/10285718.html
Copyright © 2011-2022 走看看