POJ1679 The Unique MST —— 次小生成树

题目链接：http://poj.org/problem?id=1679

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31378		Accepted: 11306

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

POJ Monthly--2004.06.27 srbga@POJ

题解：

问：最小生成树是否唯一。

次小生成树模板题。

代码如下:

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e2+;

 int cost[MAXN][MAXN], lowc[MAXN], pre[MAXN], Max[MAXN][MAXN];
 bool  vis[MAXN], used[MAXN][MAXN];

 int Prim(int st, int n)
 {
     int ret = ;
     memset(vis, false, sizeof(vis));
     memset(used, false, sizeof(used));
     memset(Max, , sizeof(Max));

     for(int i = ; i<=n; i++)
         lowc[i] = (i==st)?:INF;
     pre[st] = st;

     for(int i = ; i<=n; i++)
     {
         int k, minn = INF;
         for(int j = ; j<=n; j++)
             if(!vis[j] && minn>lowc[j])
                 minn = lowc[k=j];

         if(minn==INF) return -;  //不连通
         vis[k] = true;
         ret += minn;
         used[pre[k]][k] = used[k][pre[k]] = true;  //pre[k]-k的边加入生成树
         for(int j = ; j<=n; j++)
         {
             if(vis[j] && j!=k)  //如果遇到已经加入生成树的点，则找到两点间路径上的最大权值。
                 Max[j][k] = Max[k][j] = max(Max[j][pre[k]], lowc[k]);   //k的上一个点是pre[k]
             if(!vis[j] && lowc[j]>cost[k][j]) //否则，进行松弛操作
             {
                 lowc[j] = cost[k][j];
                 pre[j] = k;
             }
         }
     }
     return (ret==INF)?-:ret;
 }

 int SMST(int t1 ,int n)
 {
     int ret = INF;
     for(int i = ; i<=n; i++)   //用生成树之外的一条边去代替生成树内的一条边
     for(int j = i+; j<=n; j++)
     {
         if(cost[i][j]!=INF && !used[i][j])  //去掉了i-j路径上的某条边，但又把i、j直接连上，所以还是一棵生成树。
             ret = min(ret, t1+cost[i][j]-Max[i][j]);
     }
     return ret;
 }

 int main()
 {
     int T, n, m;
     scanf("%d", &T);
     while(T--)
     {

         scanf("%d%d",&n,&m);
         for(int i = ; i<=n; i++)
         for(int j = ; j<=n; j++)
             cost[i][j] = (i==j)?:INF;

         for(int i = ; i<=m; i++)
         {
             int u, v, w;
             scanf("%d%d%d", &u, &v, &w);
             cost[u][v] = cost[v][u] = w;
         }

         int t1 = Prim(, n);
         int t2 = SMST(t1, n);
         if(t1!=- && t2!=- && t1!=t2) printf("%d\n", t1);
         else  printf("Not Unique!\n");
     }
 }