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…
The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20421   Accepted: 7183 Description Given a connected undirected graph, tell if its minimum spanning tree is unique.  Definition 1 (Spanning Tree): Consider a connected, undir…
题意: 判断最小生成树是否唯一. 思路: 首先求出最小生成树,记录现在这个最小生成树上所有的边,然后通过取消其中一条边,找到这两点上其他的边形成一棵新的生成树,求其权值,通过枚举所有可能,通过这些权值看与原最小生成树的权值比较看其是否唯一.其实也可以理解成次小生成树加上最大边权的边后是否唯一. 代码: krusual: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath&…
可以依次枚举MST上的各条边并删去再求最小生成树,如果结果和第一次求的一样,那就是最小生成树不唯一. 用prim算法,时间复杂度O(n^3). #include<cstdio> #include<cstring> using namespace std; #define MAXN 111 #define INF (1<<30) struct Edge{ int u,v; }edge[MAXN]; int NE; int n,G[MAXN][MAXN]; int lowc…
Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 22335   Accepted: 7922 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G =…
The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 25203   Accepted: 8995 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undire…
The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24034   Accepted: 8535 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undire…
The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19941   Accepted: 6999 Description Given a connected undirected graph, tell if its minimum spanning tree is unique.  Definition 1 (Spanning Tree): Consider a connected, undir…
题目:poj 1679 The Unique MST 题意:给你一颗树,让你求最小生成树和次小生成树值是否相等. 分析:这个题目关键在于求解次小生成树. 方法是,依次枚举不在最小生成树上的边,然后加入到最小生成树上,然后把原树上加入了之后形成环的最长的边删去,知道一个最小的.就是次小生成树. 这些须要的都能够在求解最小生成树的时候处理出来. AC代码: #include <cstdio> #include <cstring> #include <iostream> #i…
题意:构成MST是否唯一 思路: 问最小生成树是否唯一.我们可以先用Prim找到一棵最小生成树,然后保存好MST中任意两个点i到j的这条路径中的最大边的权值Max[i][j],如果我们能找到一条边满足:他不是最小生成树中的边,并且它的权值等于Max[i][j],那么他就可以代替MST中的这条边,所以MST不唯一. 次小生成树总结 代码: #include<cmath> #include<stack> #include<queue> #include<string&…