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.
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!
题目大意：判断一个连通图的最小生成树是否唯一。
解题思路：先求出原图的最小生成树的权值，然后再求出原图的次小生成树。如果次小生成树的权值和最小生成树的权值相同，则说明原图的最小生成树不唯一。
求次小生成树的方法：先求出原图的最小生成树，然后枚举每条不在最小生成树中的边ei（u，v），先在最小生成树中删除从u 到 v 路径上的权值最大的边，然后将边ei加入其中，求得此时生成树的权值Wi ， 只要找出所有Wi中的最小值即是次小生成树的权值。
具体过程详见程序：
#include <iostream> #include<cstring> #include<algorithm> #include<string> #include<cmath> #include<vector> #include<cstdio> #define mem(a , b ) memset(a , b , sizeof(a)) using namespace std ; const int MAXN = 1e6 + 5 ; struct Edge { int from , to , w ; } e[MAXN] ; vector <int> G[105] ; vector <int> V ; int set[MAXN] ; bool vis[MAXN] ; // 标记边是否被访问 bool visd[MAXN] ; // 标记顶点是否被访问 int f[105][105] ; // 记录最小生成树中每个点对（u，v）之间路径的权值最大的边的权值 int W[105][105] ; int n , m ; int ans = 0 ; int find(int x) { int r = x ; while (r != set[r]) { r = set[r] ; } int t ; while (x != r) { t = set[x] ; set[x] = r ; x = t ; } return r ; } void chu() { int i ; for(i = 0 ; i <= n ; i ++) { set[i] = i ; G[i].clear() ; } mem(vis , 0) ; mem(f , 0) ; mem(visd , 0) ; mem(W , 0) ; V.clear() ; ans = 0 ; } bool cmp(Edge a , Edge b) { return a.w < b.w ; } void init() { scanf("%d%d" , &n , &m) ; chu() ; int i ; int a , b , c ; for(i = 0 ; i < m ; i ++) { scanf("%d%d%d" , &a , &b , &c) ; e[i].from = a ; e[i].to = b ; e[i].w = c ; } sort(e , e + m , cmp) ; int cnt = 0 ; for(i = 0 ; i < m ; i ++) { int tx , ty ; tx = e[i].from ; ty = e[i].to ; int x , y ; x = find(tx) ; y = find(ty) ; if(x != y) { G[tx].push_back(ty) ; G[ty].push_back(tx) ; W[tx][ty] = W[ty][tx] = e[i].w ; cnt ++ ; ans += e[i].w ; if(x < y) set[y] = x ; else set[x] = y ; vis[i] = true ; if(cnt == n - 1) break ; } } } void dfs(int u) // 求最小生成树中每个点对（u，v）之间的路径中权值最大的边的权值 { visd[u] = true ; V.push_back(u) ; int i ; for(i = 0 ; i < G[u].size() ; i ++) { int v = G[u][i] ; if(!visd[v]) { int j ; for(j = 0 ; j < V.size() ; j ++) { f[v][ V[j] ] = f[ V[j] ][v] = max(f[ V[j] ][u] , W[u][v]) ; } dfs(v) ; } } } void solve() { dfs(1) ; int tmp ; int i ; for(i = 0 ; i < m ; i ++) { if(!vis[i]) { int a , b ; a = e[i].from ; b = e[i].to ; tmp = ans ; tmp = ans + e[i].w - f[a][b] ; if(tmp == ans) { puts("Not Unique!") ; return ; } } } printf("%d\n" , ans) ; } int main() { int T ; scanf("%d" , &T) ; while (T --) { init() ; solve() ; } return 0 ; }