1679 The Unique MST

解题报告：The Unique MST
题目来源：pku online judge 1679
解法或类型：图论
作者：胡华嵩
The Unique MST
Time Limit:1S Memory Limit:1000K 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
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!
解题思路：
用Prim算法求MST的过程中，每一条边的增加与否只与前面过程中已扩展的点有关，而与这些点的扩展方式无关，即对于前面的点是由哪条边扩展出来的无关。

于是可以在扩展每个点的时候，判断是否有唯一的扩展方式，如果不唯一，则说明最后的MST不唯一。

由于每个点的扩展与否是根据这个点与已扩展的点集的距离决定的，每次都扩展距离最短的点。

所以在判断是否有唯一的扩展方式的时候，只需要判断一下当前需要扩展的点到已扩展的点集中每个点的距离是否有两个或两个以上的最小值，如果有，可以任意选取一条边进行扩展，即MST不唯一。

如果每步扩展的点的扩展方式都唯一，则最后的MST也唯一。

数据结构：
不用什么特殊的数据结构
时空分析：
与Prim算法时空复杂度一样
源程序：
∙#include <iostream.h>
∙#include <algorithm>
∙const int INF=1000000000;
∙
∙struct Edge{
∙ int start,stop;
∙ int weight;
∙};
∙#define N 105
∙int nn;
∙int graph[N][N];
∙Edge mst[N];
∙
∙int prim(){//用Prim算法求MST的模版，by kicc
∙ int result=0;
∙ int i,j,min,vx,vy;
∙ int weight,minweight;
∙ for(i=0;i<nn-1;i++){
∙ mst[i].start=0;
∙ mst[i].stop=i+1;
∙ mst[i].weight=graph[0][i+1];
∙ }
∙ for(i=0;i<nn-1;i++){
∙ minweight=INF;min=-1;
∙ for(j=i;j<nn-1;j++)
∙ if(mst[j].weight<minweight)
∙ minweight=mst[j].weight,min=j;
∙ if(min==-1)break;
∙ std::swap(mst[min],mst[i]);
∙ result+=minweight;
∙//下面这个循环判断每个点的扩展方式是否唯一
∙//即比较已扩展的点集中每个点到当前需要扩展的点的距离是否也等于求得的最小距离，若有，则返回-1，表示不唯一
∙for(j=0;j<i;j++)
∙if(graph[mst[j].stop][mst[i].stop]==minweig ht&&mst[j].stop!=mst[i].start)
∙return -1;
∙ vx=mst[i].stop;
∙ for(j=i+1;j<nn-1;j++){
∙ vy=mst[j].stop;
∙ weight=graph[vx][vy];
∙ if(weight<mst[j].weight)
∙ mst[j].weight=weight,mst[j].start=vx; ∙ }
∙ }
∙ return result;
∙}
∙
∙int main()
∙{
∙int d,i,j,n,m,result,x,y;
∙cin>>d;
∙while(d--)
∙{
∙cin>>n>>m;
∙nn=n;
∙for(i=0;i<n;i++)
∙for(j=0;j<n;j++)
∙graph[i][j]=INF;
∙for(i=0;i<m;i++)
∙{
∙cin>>x>>y;
∙x--;y--;
∙cin>>graph[x][y];
∙graph[y][x]=graph[x][y];
∙}
∙result=prim();
∙if(result<0) cout<<"Not Unique!"<<endl; ∙else cout<<result<<endl;
∙}
∙return 0;
∙}。