poj 1679 The Unique MST（次小生成树）

链接：

http://poj.org/problem?id=1679

题目：

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

题目大意：

问一个图的最小生成树是否是唯一存在的。

分析与总结：

一个图的最小生成树是否是唯一的，那么只需要看它的次小生成树是否和最小生成树相等，是的话就说明不是唯一的。

所以这题的关键在于求最小生成树。

在最先生成树的基础上， 添加一条不属于构成最小成树的边，那么就有了共n条边，显然会形成一个回环，所以，需要除去这个回环中的最大那条边（不包括添加的这条），那么就会形成一个新的生成树。 依次枚举添加每一条，结果最小的那个就是次小生成树。

在用prim求最小生成树时，用一个二位数组path【a】【b】, 来存a到b的最长一条边。

之前自己整了个Prim的模板，做这题只稍微改了下模板。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

const int VN = 105; //尺寸
const int INF = 2000000000;
   
template<typename Type>
class Prim{
public:
    inline void init(int _n){
        n=_n; MSTsum = 0;
        for(int i=0; i<=n; ++i){
            w[i][i] = INF;
            for(int j=i+1; j<=n; ++j)
                w[i][j]=w[j][i]=INF;
        }
    }
    inline void insert(int u, int v, Type weight){
        if(w[u][v]>weight) w[u][v] = weight; //注意可能有重复边
    }
    Type prim(){
        memset(vis, 0, sizeof(vis));
        memset(path, 0, sizeof(path));
        memset(used, 0, sizeof(used));
        vis[1] = true;
        for(int i=1; i<=n; ++i){
            key[i] = w[1][i]; pre[i] = 1; 
        }
        for(int i=1; i<n; ++i){
            int u=-1;
            for(int j=1; j<=n; ++j)if(!vis[j]){
                if(u==-1||key[j]<key[u]) u=j;
            }
            if(w[pre[u]][u]==INF) return -1; 
            used[pre[u]][u] = used[u][pre[u]] = true;
            MSTsum += w[pre[u]][u];
            vis[u] = true;
            for(int j=1; j<=n; ++j){
                if(vis[j]&&j!=u){
                    path[u][j]=path[j][u]=max(path[j][pre[u]],key[u]);
                }
                if(!vis[j]&&key[j]>w[u][j]){
                    key[j] = w[u][j]; pre[j] = u;
                }
            }
        }
        return MSTsum;
    }
    int is_unique(){
        prim();
        int tmp;
        for(int i=1; i<=n; ++i){
            for(int j=i+1; j<=n; ++j)if(!used[i][j]){
                tmp = MSTsum + path[i][j] - w[i][j];
                if(tmp==MSTsum) return -1;
            }
        }
        return MSTsum;
    }
   
private:
    Type MSTsum;
    Type w[VN][VN], key[VN];
    int pre[VN], path[VN][VN], n;
    bool vis[VN], used[VN][VN];
};
Prim<int>G;

int main(){
    int T, n, m, a, b, c;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        G.init(n);
        for(int i=0; i<m; ++i){
            scanf("%d%d%d",&a,&b,&c);
            G.insert(a,b,c);
            G.insert(b,a,c);
        }
        int ans = G.is_unique();
        if(ans==-1) puts("Not Unique!");
        else printf("%d\n", ans);
    }
    return 0;
}

——  生命的意义，在于赋予它意义。
 
  

             
  
  

        原创 http://blog.youkuaiyun.com/shuangde800 ， By   D_Double  (转载请标明)