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 following properties: 
1. V' = V. 
2. T is connected and acyclic.

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

Sample Input

2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

Sample Output

3
Not Unique!
方法1:首先算出最小生成树的权值和ans,然后枚举删除最小生成树中的每一条边,若还可以达到相同的效果,就说明最小生成树不唯一,
因为两个不同的最小生成树至少有一条边不同,所以我们才可以枚举删除每一条边.
方法2:判断最小生成树和次小生成树的权值是否相同.
#include<iostream>
#include<vector>
#include<algorithm>
#include<stdio.h>
using namespace std;
typedef long long ll;
const int maxn=;
int f[maxn];
struct node
{
int u,v,w;
bool operator < (const node &r)const{
return w<r.w;
}
}q[maxn];
int Find(int x)
{
return f[x]==x?x:f[x]=Find(f[x]);
}
int Merge(int u,int v)
{
u=Find(u);
v=Find(v);
if(u!=v)return f[u]=v,;
return ;
}
vector<int>v;
int main()
{
int T;
cin>>T;
while(T--){
v.clear();
int n,m;
cin>>n>>m;
for(int i=;i<=n;i++)f[i]=i;
for(int i=;i<=m;i++){
cin>>q[i].u>>q[i].v>>q[i].w;
}
sort(q+,q++m);
int ans=;
for(int i=;i<=m;i++){
int x=Merge(q[i].u,q[i].v);
if(x){
v.push_back(i);
ans+=q[i].w;
}
}
int flag=;
for(int i=;i<v.size();i++){
int sum=,cnt=;
for(int j=;j<=n;j++)f[j]=j;
for(int j=;j<=m;j++){
if(j==v[i])continue;
int x=Merge(q[j].u,q[j].v);
if(x){
sum+=q[j].w;
cnt++;
}
}
if(cnt==n-&&ans==sum){
flag=;
break;
}
}
if(flag)cout<<ans<<endl;
else printf("Not Unique!\n"); }
return ;
}
#include<iostream>
#include<cstring> using namespace std;
typedef long long ll;
const int maxn=;
const int INF=0x3f3f3f3f;
int Maxlen[maxn][maxn];
int dis[maxn],vis[maxn];
int pre[maxn],MAP[maxn][maxn];
int used[maxn][maxn];
int n,m; int Prim(int x)
{
memset(Maxlen,,sizeof(Maxlen));
memset(dis,INF,sizeof(dis));
memset(vis,,sizeof(vis));
memset(pre,,sizeof(pre));
memset(used,,sizeof(used));
for(int i=;i<=n;i++){
dis[i]=MAP[x][i];
pre[i]=x;
}
dis[x]=;
vis[x]=;
pre[x]=;
int ans=;
for(int i=;i<=n;i++){
int u=,minn=INF;
for(int j=;j<=n;j++){
if(!vis[j]&&dis[j]<minn){
u=j;
minn=dis[j];
}
}
vis[u]=;
ans+=minn;
used[u][pre[u]]=used[pre[u]][u]=;
for(int v=;v<=n;v++){
if(vis[v]){
Maxlen[u][v]=Maxlen[v][u]=max(Maxlen[v][pre[u]],dis[u]);
}
else{
if(dis[v]>MAP[u][v]){
dis[v]=MAP[u][v];
pre[v]=u;
}
}
}
}
return ans;
}
void sst(int ans)
{
int sum=INF;
for(int i=;i<=n;i++){
for(int j=i+;j<=n;j++){
if(!used[i][j]&&MAP[i][j]!=INF){
sum=min(sum,ans+MAP[i][j]-Maxlen[i][j]);
}
}
}
if(sum==ans)cout<<"Not Unique!"<<endl;
else cout<<ans<<endl;
}
int main()
{
int T;
cin>>T;
while(T--){
cin>>n>>m;
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
if(i==j)MAP[i][j]=;
else MAP[i][j]=INF;
}
}
for(int i=;i<=m;i++){
int u,v,w;
cin>>u>>v>>w;
MAP[u][v]=MAP[v][u]=min(MAP[u][v],w);
}
int ans=Prim();
sst(ans);
}
return ;
}

K - The Unique MST (最小生成树的唯一性)的更多相关文章

  1. The Unique MST(最小生成树的唯一性判断)

    Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spann ...

  2. K - The Unique MST

    K - The Unique MST #include<iostream> #include<cstdio> #include<cstring> #include& ...

  3. [poj1679]The Unique MST(最小生成树)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28207   Accepted: 10073 ...

  4. POJ 1679 The Unique MST (最小生成树)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 22668   Accepted: 8038 D ...

  5. K - The Unique MST - poj 1679

    题目的意思已经说明了一切,次小生成树... ****************************************************************************** ...

  6. POJ1679 The Unique MST(Kruskal)(最小生成树的唯一性)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 27141   Accepted: 9712 D ...

  7. poj1679 The Unique MST(最小生成树唯一性)

    最小生成树的唯一性,部分参考了oi-wiki 如果一条不在最小生成树边集内的边,它可以替换一条在最小生成树边集内,且权值相等的边,那么最小生成树不是唯一的 同过kruskal来判断 考虑权值相等的边, ...

  8. (poj)1679 The Unique MST 求最小生成树是否唯一 (求次小生成树与最小生成树是否一样)

    Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definit ...

  9. The Unique MST (判断是否存在多个最小生成树)

    The Unique MST                                                                        Time Limit: 10 ...

随机推荐

  1. A. Yellow Cards ( Codeforces Round #585 (Div. 2) 思维水题

    ---恢复内容开始--- output standard output The final match of the Berland Football Cup has been held recent ...

  2. python安装wordcloud、jieba,pyecharts

    1.安装wordcloud: 适用于无法使用pip install wordcloud安装的情况: 据python和windows 版本 到https://www.lfd.uci.edu/~gohlk ...

  3. [转]java 的HashMap底层数据结构

    java 的HashMap底层数据结构   HashMap也是我们使用非常多的Collection,它是基于哈希表的 Map 接口的实现,以key-value的形式存在.在HashMap中,key-v ...

  4. 磁盘报No space left on device,但是 df -h 查看磁盘空间没满

    df -h Filesystem Size Used Avail Use% Mounted on /dev/mapper/dev01-root 75G 58G 14G 82% / udev 2.0G ...

  5. for循环和增强for循环

  6. 十二星座 英文名:Aries 金牛座 (4/21 - 5/20)的英文名: Taurus 双子座 (5/21 - 6/21)的英文名: Gemini 巨蟹座 (6/22 - 7/22)的英文名: Cancer 狮子座 (7/23 - 8/22)的英文名: Leo 处女座/室女座 (8/23 - 9/22)的英文名: Virgo 天秤座 (9/2

    十二星座的具体顺序是:白羊座(Aries).金牛座(Taurus).双子座(Gemini).巨蟹座(Cancer).狮子座(Leo).处女座(Virgo).天秤座(Libra).天蝎座(Scorpio ...

  7. Paper Review: Epigenetic Landscape, Cell Differentiation 02

    I'll share another review paper about Epigenetic Landscape, it comes from Nature Review, published i ...

  8. android studio 修改新建EmptyActivity默认布局

    https://www.jianshu.com/p/d4f201135097 打开你的Android Sudio安装目录,我的为D:\Program Files\Android\Android Stu ...

  9. python编程:从入门到实践----第五章>if 语句

    一.一个简单示例 假设有一个汽车列表,并想将其每辆汽车的名称打印出来.遇到汽车名‘bmw’,以全大写打印:其他汽车名,首字母大写 cars=['audi','bmw','subaru','toyota ...

  10. Window RabbitMq安装

    rabbitMQ是一个在AMQP协议标准基础上完整的,可服用的企业消息系统.它遵循Mozilla Public License开源协议,采用 Erlang 实现的工业级的消息队列(MQ)服务器,Rab ...