K - The Unique MST (最小生成树的唯一性)
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
Output
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 ;
}
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