poj 1679 The Unique MST(最小树不唯一的判定)

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16068		Accepted: 5566

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 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!

Source

POJ Monthly--2004.06.27 srbga@POJ

思路：按并查集，先解出第一次的最小树，然后标记所有用过的边，和所具有相同权值的边。然后依次删除用过的具有相同权值的边，看最小生成树的值是不是一样，一样则最             小树不唯一。

易错点：权值全为0 要注意。

               2 

               3 2

               1 2 0

               2 3 0

               3 3

                1 2 0

                 2 3 0

                1 3 0

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int mm=110000;
const int oo=1e9;
int root[mm],ran[mm];
class node
{
  public:int u,v,c;
  bool used,equ,del;
  node(){used=equ=del=0;}
}edge[mm];
int m,n,cen;
bool cmp(node a,node b)
{
  return a.c<b.c;
}
int look(int x)
{
  if(x^root[x])root[x]=look(root[x]);
  return root[x];
}
void uni(int x,int y)
{
  if(ran[look(x)]>ran[look(y)])root[look(y)]=look(x),ran[look(x)]+=ran[look(y)];
  else root[look(x)]=look(y),ran[look(y)]+=ran[look(x)];
}
int kruskal()
{ cen++;int sum=0,num=0;
  for(int i=0;i<m;i++)
  {
    if(edge[i].del)continue;
    if(look(edge[i].u)^look(edge[i].v))
    {uni(edge[i].u,edge[i].v);
     sum+=edge[i].c;num++;
     if(cen==1)
     edge[i].used=1;
    }
    if(num>=n-1)break;
  }
  if(num<n-1)return -oo;///不能构成最小树
  return sum;
}
void _set()
{
  for(int i=0;i<=n;i++)
      root[i]=i,ran[i]=1;
}
int main()
{
  int cas;
  while(cin>>cas)
  { while(cas--)
    { memset(edge,0,sizeof(edge));
      cin>>n>>m;
      for(int i=0;i<m;i++)
      {
      cin>>edge[i].u>>edge[i].v>>edge[i].c;
      }
      for(int i=0;i<m;i++)
      for(int j=0;j<m;j++)
      if(i^j&&edge[i].c==edge[j].c)
      edge[i].equ=edge[j].equ=1;
      sort(edge,edge+m,cmp);
     _set();int ans=0,z;
     cen=0;ans=kruskal();
     ///if(ans==-oo)ans=0;
     bool flag=1;
     for(int i=0;i<m;i++)
     {
       if(edge[i].used&&edge[i].equ)
       {  _set();
         edge[i].del=1;z=kruskal();edge[i].del=0;
         if(z==ans){flag=0;break;}
       }
     }
     if(!flag)cout<<"Not Unique!\n";
     else cout<<ans<<"\n";
   }
  }
}