POJ 1679 The Unique MST【MST是否唯一，Prime算法，最好的代码】

The Unique MST Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27389 Accepted: 9816

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

原题链接：http://poj.org/problem?id=1679

题意：看最小生成树是否唯一，网上看了好多的代码，都用到了枚举，但是发现了一份不用枚举的代码，

原文链接：http://blog.csdn.net/cambridgeacm/article/details/7857252，但是有一点不明白，就是主循环为什么是n次，而不是n-1次。有大神知道吗？

AC代码：

#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int INF=0x3f3f3f3f; int a[105][105]; int dis[105]; bool vis[105]; int n,m; void Prime() { for(int i=1; i<=n; i++) { dis[i]=a[1][i]; vis[i]=false; } int ans=0; bool flag=false; for(int i=1; i<=n; i++) { int minn=INF; int p=-1; for(int j=1; j<=n; j++) { if(!vis[j]&&dis[j]<minn) minn=dis[p=j]; } int k=0; for(int j=1; j<=n; j++) { if(vis[j]&&a[p][j]==minn) k++; } if(k>1) { flag=true; break; } vis[p]=1; ans+=minn; for(int j=1; j<=n; j++) { if(!vis[j]&&dis[j]>a[p][j]) dis[j]=a[p][j]; } } if(flag) cout<<"Not Unique!"<<endl; else cout<<ans<<endl; } int main() { int T; freopen("data/1679.txt","r",stdin); cin>>T; while(T--) { cin>>n>>m; for(int i=0; i<=n; i++) { for(int j=0; j<=n; j++) if(i==j) a[i][j]=0; else a[i][j]=INF; } int x,y,z; while(m--) { scanf("%d%d%d",&x,&y,&z); if(z<a[x][y]) a[x][y]=a[y][x]=z; } Prime(); } return 0; }