判定最小生成树是否唯一

题目链接:http://poj.org/problem?id=1679 The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 29408   Accepted: 10520 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spannin

The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21931   Accepted: 7784 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undire

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 22346   Accepted: 7924 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: 20679   Accepted: 7255 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undire

/* 只需判断等效边和必选边的个数和n-1的关系即可 */ #include<stdio.h> #include<stdlib.h> #define N 110 struct node { int u,v,w; }f[N*N*2]; int cmp(const void *a,const void*b) { return (*(struct node *)a).w-(*(struct node *)b).w; } int pre[N]; int find(int x) { if(x

The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20430   Accepted: 7186 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undire

设G是所有边权均不相同的无向联通图. 证明一: 首先,易证图G中权值最小的边一定是最小生成树中的边.(否则最小生成树加上权值最小的边后构成一个环,去掉环中任意一条非此边则形成了另一个权值更小的生成树). 之后用反证法,假设G存在俩个不同的最小生成树 ①.设G的俩个不同的最小生成树T1 T2,设这俩颗生成树的并集为子图G1,G1为连通图且T1 T2显然为G1的最小生成树,由首先可得知俩颗生成树至少包含一条公共边,将G1中两颗生成树的公共边删去,得到子图G2.G2由一个或多个连通分量组成,其中至少有

The Unique MST Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21646 Accepted: 7661 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected

链接:poj 1679 题意:判断最小生成树是否唯一, 若唯一,输出最小权值和,否则,输出  Not Unique! 判断最小生成树是否唯一的思路: 1.对图中的每一条边,扫描其他边,如果存在相同权值的边,则对该边做标记 2.然后用Kruskal算法或Prim算法求MST 3.求得MST后,如果该MST中未包含做了标记的边,即可判断MST唯一: 如果包含做了标记的边,则依次去掉这些边的一条边,再求MST, 如果求得的MST权值和原来的MST的权值一样,即可判断MST不唯一. 个人思路是先求最小生