The Unique MST （判断是否存在多个最小生成树）

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　The Unique MST                                                                       
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 24058   Accepted: 8546

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!
题解：
这个题看有的人还求了次小生成树，其实不必要的，中间判断下是否存在多个生成树就好了；
代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 const int INF=0x3f3f3f3f;
 4 const int MAXN=110;
 5 int vis[MAXN],map[MAXN][MAXN],low[MAXN];
 6 int ans,N,answer;
 7 void prime(){
 8     memset(vis,0,sizeof(vis));
 9     int flot=1,temp,k;
10     answer=0;
11     ans=1;
12     vis[1]=1;
13     for(int i=1;i<=N;i++)low[i]=map[1][i];
14     for(int i=1;i<=N;i++){
15         temp=INF;
16         for(int j=1;j<=N;j++)
17         if(!vis[j]&&temp>low[j])temp=low[k=j];
18         if(temp==INF){
19             if(flot!=N)ans=0;//因为这题本来就是联通的，所以这句话可以省略； 
20         //    printf("flot=%d,N=%d\n",flot,N);
21             break;
22         }
23         int x=0;
24         for(int j=1;j<=N;j++){
25             if(vis[j]&&map[k][j]==temp)x++;
26             if(x>1)break;
27         }//这个for循环的作用就是，当前已经找到距离这个图最小的点了，然后判断到这个最小点的距离为temp也就是最小值的点有几个，如果点数大于一则有多个最小生成树； 
28         if(x>1){
29             ans=0;
30             break;
31         }
32         answer+=temp;
33         vis[k]=1;
34         flot++;
35         for(int j=1;j<=N;j++){
36             if(!vis[j]&&low[j]>map[k][j])low[j]=map[k][j];
37             }
38         }
39     }
40 int main(){
41                 int T,M,a,b,c;
42                 scanf("%d",&T);
43         while(T--){
44             memset(map,INF,sizeof(map));
45                 scanf("%d%d",&N,&M);
46                 while(M--){
47                     scanf("%d%d%d",&a,&b,&c);
48                     if(c<map[a][b])map[a][b]=map[b][a]=c;
49                 }
50                 prime();
51                 if(ans)printf("%d\n",answer);
52                 else printf("Not Unique!\n");
53                     }    
54             return 0;
55 }

 

posted @ 2015-08-13 10:52  handsomecui  阅读(733)  评论(0)  编辑  收藏  举报