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:
V’ = V.
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!
判断最小生成树是否唯一，这里先把最小生成树求出来并记录加入树的那些边，之后每次删去其中一条边再求最小生成树，但是删去原树的边之后有可能无法连通，需要特判。最后比较两次的代价，不相等或者无法连通则树唯一，反之则不唯一。
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define maxn 105
int fa[maxn], n, t, m, ans;
bool flag = 0;
struct Edge
{
int x, y, z;
bool flag;
} edge[maxn*maxn];
bool cmp(Edge a,Edge b)
{
return a.z < b.z;
}
int find(int x)
{
if(x==fa[x])
return x;
return fa[x] = find(fa[x]);
}
int Kruskal(int k) //删去原树中某边后可能无法构成树
{
int sum = 0;
int cnt = 1;
for (int i = 1; i <= n;i++)
fa[i] = i;
for (int i = 1; i <= m;i++)
{
if(i==k)
continue;
int x = find(edge[i].x);
int y = find(edge[i].y);
if(x==y)
continue;
fa[x] = y;
cnt++;
sum += edge[i].z;
}
if(cnt!=n) //不连通
return -1;
return sum;
}
int main()
{
cin >> t;
while(t--)
{
flag = 0;
cin >> n >> m;
for (int i = 1; i <= m;i++)
{
cin >> edge[i].x >> edge[i].y >> edge[i].z;
edge[i].flag = 0;
}
for (int i = 1; i <= n;i++)
fa[i] = i;
sort(edge + 1, edge + 1 + m, cmp);
int cnt = 0;
for (int i = 1; i <= m;i++)
{
int x = find(edge[i].x);
int y = find(edge[i].y);
if(x==y)
continue;
fa[x] = y;
edge[i].flag = 1; //标记这条边加入了MST
ans += edge[i].z;
}
for (int i = 1; i <= m;i++)
{
if(edge[i].flag==0) //跳过不在树中的边
continue;
int cnt = Kruskal(i);
if(cnt==ans)
{
flag = 1;
break;
}
}
if(flag)
cout << "Not Unique!" << endl;
else
cout << ans << endl;
ans = 0;
}
}