# [poj1679]The Unique MST(最小生成树)

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

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

http://www.docin.com/p-806495282.html
 1 #include<cstdio>
2 #include<cstdlib>
3 #include<cstring>
4 #include<cmath>
5 #include<climits>
6 #include<algorithm>
7 #include<queue>
8 #define LL long long
9 using namespace std;
10 typedef struct{
11     int to,frm,dis;
12 }edge;
13 edge gra[20010];
14 int num=0,fa[110];
15 int n,m;
16 int cmp(const edge &a,const edge &b){
17     return a.dis<b.dis;
18 }
19 int fnd(int x){
20     return x==fa[x]?x:fnd(fa[x]);
21 }
22 int uni(int x,int y){
23     int fx=fnd(x);
24     int fy=fnd(y);
25     fa[fy]=fx;
26     return 0;
27 }
29        int sum=0;char ch=getchar();
30        while(ch>'9'||ch<'0')ch=getchar();
31        while(ch<='9'&&ch>='0'){
32              sum=sum*10+ch-'0';
33              ch=getchar();
34        }
35        return sum;
36 }
37 int kru(){
38     int ans=0;
39     sort(gra+1,gra+m+1,cmp);
40     for(int i=1;i<=n;i++)fa[i]=i;
41     for(int i=1;i<=m;i++){
42         int x=gra[i].frm;
43         int y=gra[i].to;
44         int fx=fnd(x);
45         int fy=fnd(y);
46         if(fx!=fy){
47             int j=i+1;
48             while(j<=m&&gra[j].dis==gra[i].dis){
49                 int y1=gra[j].frm;
50                 int x1=gra[j].to;
51                 int fy1=fnd(y1);
52                 int fx1=fnd(x1);
53                 if((fx1==fx&&fy1==fy)||(fx1==fy&&fy1==fx))return -1;
54                 j++;
55             }
56             ans+=gra[i].dis;
57             uni(fx,fy);
58         }
59     }
60     return ans;
61 }
62 int main(){
63     int t;
65     while(t--){
66         memset(gra,0,sizeof(gra));
68           num=0;
69         for(int i=1;i<=m;i++){
80 }