Problem 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!題目意思是：在給定的數據中，找到一棵最小生成樹，如果這棵最小生成樹是唯一的那麼就輸出最小生成樹上權的和，不是唯一的就輸出Not Unique!解法：先建立一棵最小生成樹，然後找一條邊不在最小生成樹上，把這條邊作為起始邊開始建最小生成樹，如果找到一棵最小生成樹與第一次建的最小生成樹權值相等，那麼就不用再找了，直接輸出Not Unique!
#include<stdio.h>
#include<string.h>
typedef struct sn {
int v;//用於記錄與當前點相連的點
int dist;//權值 }Node;
Node node[105];
int n,s[105],vist[105][105],map[105][105],INF=1000000;
void set_first(int n)//初始化
{ for(int i=1;i<=n;i++)
{ s[i]=0; node[i].dist=INF;
for(int j=i+1;j<=n;j++) map[i][j]=map[j][i]=INF;
} memset(vist,0,sizeof(vist));
} int Prim(int m,int sum,int flog) { int tm=m,min,i;
s[m]=1;
if(flog==0)i=2;//如果flog=0就是s集合己有1個點，否則有2個點 else i=3;
for( ;i<=n;i++) { min=INF;
for(int j=1;
j<=n;j++) if(s[j]==0) {
if(node[j].dist>map[tm][j])//更新權值，並且更新與j相連的點tm { node[j].dist=map[tm][j];
if(flog==0) node[j].v=tm;
} if(min>node[j].dist)//找到最小的權值並記錄點位置 { min=node[j].dist; m=j;
} } sum+=min; tm=m; s[m]=1;
if(flog==0)
vist[node[m].v][m]=vist[m][node[m].v]=1;
//為1時邊在最小生成樹上，否則不在樹上 } return sum;
} int main() { int t,a,b,w,m,sum1,sum2;
scanf("%d",&t);
while(t--) { scanf("%d%d",&n,&m);
set_first(n);//初始化
while(m--) { scanf("%d%d%d",&a,&b,&w);
if(map[a][b]>w)//判斷重邊
map[a][b]=map[b][a]=w; } sum1=Prim(1,0,0);
int flog=0;
for(int i=1;i<n;i++)//找一條不在最小生成樹上的邊，然後從這條邊開始建樹 { for(int j=i+1;j<=n;j++)//從i+1開始為了不重復，因為如：1到2和2到1是一樣的
if(vist[i][j]==0&&map[i][j]!=INF)
{ for(int e=1;e<=n;e++)//初始化 {
s[e]=0; node[e].dist=INF;
}
s[i]=1;//加入第一個點 for(int e=1;e<=n;e++)//更新與第一個點相關的點 { node[e].dist=map[i][e];
} sum2=Prim(j,node[j].dist,1);//從點j開始找 if(sum2==sum1){flog=1;break;}//找到有多種可能，跳出
} if(flog)break; } if(flog==0)
printf("%d\n",sum1);
else printf("Not Unique!\n"); } }
