POJ 1679 The Unique MST
The Unique MST
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 29079 | Accepted: 10398 |
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'.
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
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 const int VM=120; 7 const int EM=10010; 8 struct Edge{ 9 int u,v,cap; 10 bool operator < (const Edge& a)const { 11 return cap<a.cap; 12 } 13 }edge[EM<<1]; 14 int n,m,flag,ans,father[VM]; 15 void makeSet(){ 16 for(int i=1;i<=n;i++) 17 father[i]=i; 18 } 19 int findSet(int x){ 20 if(x==father[x]) return x; 21 else return father[x]=findSet(father[x]); 22 } 23 void Kruskal(){ 24 makeSet(); 25 sort(edge,edge+m); 26 int path[EM],cnt=0; 27 ans=0; 28 for(int i=0;i<m;i++){ 29 int u=findSet(edge[i].u); 30 int v=findSet(edge[i].v); 31 if(u!=v){ 32 father[v]=u; ans+=edge[i].cap; 33 path[cnt++]=i; //记录路径 34 } 35 } 36 for(int k=0;k<cnt;k++){ //枚举去掉每一条边 37 makeSet(); 38 int sum=0,j=0; 39 for(int i=0;i<m;i++){ 40 if(i==path[k]) continue; 41 int u=findSet(edge[i].u),v=findSet(edge[i].v); 42 if(u!=v){ 43 father[v]=u; 44 sum+=edge[i].cap; 45 j++; 46 } 47 } 48 if(j==n-1 && sum==ans){ //判断是否能构成树 且 是否与最小生成树相等 49 flag=0;return ; 50 } 51 } 52 } 53 int main(){ 54 int T; 55 scanf("%d",&T); 56 while(T--){ 57 scanf("%d%d",&n,&m); 58 for(int i=0;i<m;i++) 59 scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].cap); 60 flag=1; 61 Kruskal(); 62 if(flag) printf("%d\n",ans); 63 else printf("Not Unique!\n"); 64 } 65 return 0; 66 }
思路：先来一遍最小生成树，此时顺便记录下树边，然后枚举删除掉树边，每次删除后看是否还能生成一棵最小生成树和原最小生成树的MST相等..
