-*- 父页面js function mapFocus(){ //console.log("-*-"); var longitude = mini.get("jd").getValue(); // 经度 var latitude = mini.get("wd").getValue(); // 纬度 var url = "<%=basePath %>project/construction/Map.jsp"; //
判断值是否在set集合中的速度明显要比list快的多, 因为查找set用到了hash,时间在O(1)级别. 假设listA有100w个元素,setA=set(listA)即setA为listA转换之后的集合. 以下做个简单的对比: for i in xrange(0, 5000000): if i in listA: pass for i in xrange(0, 5000000): if i in setA: pass 第一个循环用了16min,第二个循环用了52s. 由此可见,在set中判断
题目链接 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the fol
http://poj.org/problem?id=1679 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say
题意:给出n个字符,m对关系,让你输出三种情况: 1.若到第k行时,能判断出唯一的拓扑序列,则输出: Sorted sequence determined after k relations: 序列 2.若到第k行,出现环,则输出: Inconsistency found after k relations. 3.若直到m行后,仍判断不出唯一的拓扑序列,则输出: Sorted sequence cannot be deter
题目链接: http://poj.org/problem?id=1679 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G
题意:给出n个点,m条边,要你判断最小生成树是否唯一. 思路:先做一次最小生成树操作,标记选择了的边,然后枚举已经被标记了的边,判断剩下的边组成的最小生成树是否与前面的相等,相等,则不唯一,否则唯一...... #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct node { int v1,v2; int d