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1.先生成最小生成树，长度为MinLen，并求出点i到点j的最大边(Max[i][j]).2.再枚举每一条不在树上的边，将该边L[i][j]添加到树中，此时构成了一个环，然后删除该环上(除L[i][j]外)的最大权值边Max[i][j],生成了一颗新的树，len = MinLen + L[i][j] - Max[i][j] ，再算出所有新树中的最小len，即为次小生成树的最小len。例

题目链接：传送门The Unique MSTTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 31839 Accepted: 11504DescriptionGiven a connected undirected graph, tell i

题目链接：传送门1965: MessageSubmit Page    Summary    Time Limit: 1 Sec     Memory Limit: 128 Mb     Submitted: 68     Solved: 7    DescriptionYou are a boss who has N employe

题目链接：传送门Balancing ActTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14208 Accepted: 6008DescriptionConsider a tree T with N (1 <= N <= 20,000)

题目链接：传送门Park VisitTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3743    Accepted Submission(s): 1679Problem DescriptionC

题目链接： 传送门题目大意：给出一个n(2 &lt;= n &lt;= 100)个结点的无向图，找一颗苗条度（最大边减最小边）最小的生成树。图中不包含自环和重边。解题思路：枚举树上的最小边，从这条最小边开始构建MST，此时这棵树是以该边为最小边的苗条度最小的生成树。因为构建MST时会使最大边最小。代码：#include &lt;iostream&gt;#include &lt;stack&gt;...

题目链接：传送门DescriptionICPC (Isles of Coral Park City) consist of several beautiful islands.The citizens requested construction of bridges between islands to resolve inconveniences of using boats between ...