http://www.cnblogs.com/kuangbin/p/3147329.html #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=110; const int INF=0x3f3f3f3f; bool vis[N]; int lowc[N],pre[N],Max[N][N]; bool used[N][N]; int Prim(in…

题目链接: https://vjudge.net/problem/POJ-1679 题目大意: 给定一个无向连通网,判断最小生成树是否唯一. 思路: (1)对图中的每条边,扫描其他边,如果存在相同权值的边,对该边做标记. (2)然后用kruskal算法或者prim算法求MST(标记MST中的边) (3)求得MST后,如果MST中未包含做了标记的边,那么MST唯一. MST中不包含未标记的边,说明MST中没有那些权值相同的边,用kruskal算法可知,该最小生成树肯定唯一. (4)如果包含标记的边…

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…

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…

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 followi…

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…

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…

借用的是Kruskal的并查集,算法中的一点添加和改动. 通过判定其中有多少条可选的边,然后跟最小生成树所需边做比较,可选的边多于所选边,那么肯定方案不唯一. 如果不知道这个最小生成树的算法,还是先去理解下这个算法的原理,再继续看. 多加的几行与cnt2很类似的cnt1统计,是统计当前未选边中相同长度的边还有哪些可以选,但不标记,只是为了把当前可选的边全部统计出来. 其他细节,自己要细心点- #include <stdio.h> #include <string.h> #inclu…

Kruskal模板:按照边权排序,开始从最小边生成树 #include<algorithm> #include<stdio.h> #include<string.h> #include<iostream> #define N 1000+5//n 个顶点,m条边 using namespace std; //最小生成树模板(计算最小生成树的sum) struct node { int u,v,len;//u->v距离len }q[N]; int f[N]…