http://poj.org/problem?id=1679
The Unique MST
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 17726 | Accepted: 6150 |
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'.
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
【题解】：
用kruskal，求出最小生成树，然后枚举删除最小生成树上的一条边，再求最小生成树，比较前后两次求解的大小
【注意】：如果题目给的数据不能构成一个树需要输出0
【code】：
1 /** 2 Judge Status:Accepted Memory:748K 3 Time:32MS Language:G++ 4 Code Lenght:1814B Author:cj 5 */ 6 #include<iostream> 7 #include<stdio.h> 8 #include<string.h> 9 #include<algorithm> 10 #include<vector> 11 12 #define N 101 13 #define M 6000 14 using namespace std; 15 16 struct Nod 17 { 18 int x,y,w; 19 }node[M]; 20 21 int parent[N]; 22 int n,m; 23 vector<int> vct; 24 25 bool cmp(Nod a,Nod b) 26 { 27 return a.w<b.w; 28 } 29 30 int findp(int a) 31 { 32 while(a!=parent[a]) 33 { 34 a=parent[a]; 35 } 36 return a; 37 } 38 39 int merge(Nod nd) //合并 40 { 41 int x = findp(nd.x); 42 int y = findp(nd.y); 43 if(x>y) 44 { 45 parent[y]=x; 46 return nd.w; 47 } 48 else if(x<y) 49 { 50 parent[x]=y; 51 return nd.w; 52 } 53 return -1; 54 } 55 56 int kruskal(int id) 57 { 58 int i,sum=0,cnt=0; 59 for(i=1;i<=n;i++) parent[i]=i; 60 for(i=0;i<m;i++) 61 { 62 if(i!=id) 63 { 64 int temp = merge(node[i]); 65 if(temp!=-1) 66 { 67 sum+=temp; 68 cnt++; //剪枝 69 if(id==-1) vct.push_back(i); //保存第一次最小生成树的各个节点 70 } 71 if(cnt>=n-1) //找到n-1条边即可以跳出了 72 break; 73 } 74 } 75 cnt = 0; 76 for(i=1;i<=n;i++) //判断是不是构成一棵树 77 if(parent[i]==i) 78 cnt++; 79 if(cnt==1) //是 80 return sum; 81 if(id==-1) //否 82 return 0; 83 return -1; 84 } 85 86 int main() 87 { 88 int t; 89 scanf("%d",&t); 90 while(t--) 91 { 92 scanf("%d%d",&n,&m); 93 int i; 94 for(i=0;i<m;i++) 95 { 96 scanf("%d%d%d",&node[i].x,&node[i].y,&node[i].w); 97 } 98 vct.clear(); 99 sort(node,node+m,cmp); 100 int mins = kruskal(-1); //找到第一颗最小生成树 101 int temp=-1; 102 for(i=0;i<vct.size();i++) 103 { 104 temp = kruskal(vct[i]); //每次去掉一个节点 再判断是否可以组成最小生成树 105 if(mins==temp) 106 break; 107 } 108 if(temp==mins) puts("Not Unique!"); 109 else printf("%d ",mins); 110 } 111 return 0; 112 }