POJ 1679 The Unique MST (最小生成树)

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22668		Accepted: 8038

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

Sample Output

3
Not Unique!



POJ的网站绝对有问题。昨天就有一题提交不了，换到HUD上就A了，今天这题同样没法提交，一提交就卡死，换到百炼上成功AC。
根据概念来做，如果在某一步合并的时候有多个可以选，那么就不唯一了。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <string>
  4 #include <queue>
  5 #include <vector>
  6 #include <map>
  7 #include <algorithm>
  8 #include <cstring>
  9 #include <cctype>
 10 #include <cstdlib>
 11 #include <cmath>
 12 #include <ctime>
 13 using    namespace    std;
 14 
 15 const    int    SIZE = 105;
 16 int    FATHER[SIZE],N,M,NUM;
 17 int    MAP[SIZE][SIZE];
 18 struct    Node
 19 {
 20     int    from,to,cost;
 21 }G[SIZE * SIZE];
 22 
 23 void    ini(void);
 24 int    find_father(int);
 25 void    unite(int,int);
 26 bool    same(int,int);
 27 int    kruskal(void);
 28 bool    comp(const Node &,const Node &);
 29 int    main(void)
 30 {
 31     int    t;
 32     scanf("%d",&t);
 33     while(t --)
 34     {
 35         scanf("%d%d",&N,&M);
 36         ini();
 37         for(int i = 0;i < M;i ++)
 38         {
 39             scanf("%d%d%d",&G[NUM].from,&G[NUM].to,&G[NUM].cost);
 40             NUM ++;
 41         }
　　　　　　　sort(G,G＋NUM,comp);
 42         kruskal();
 43     }
 44 
 45     return 0;
 46 }
 47 
 48 void    ini(void)
 49 {
 50     NUM = 0;
 51     for(int i = 1;i <= N;i ++)
 52         FATHER[i] = i;
 53 }
 54 
 55 int    find_father(int n)
 56 {
 57     if(FATHER[n] == n)
 58         return    n;
 59     return    FATHER[n] = find_father(FATHER[n]);
 60 }
 61 
 62 void    unite(int x,int y)
 63 {
 64     x = find_father(x);
 65     y = find_father(y);
 66 
 67     if(x == y)
 68         return    ;
 69     FATHER[x] = y;
 70 }
 71 
 72 bool    same(int x,int y)
 73 {
 74     return    find_father(x) == find_father(y);
 75 }
 76 
 77 bool    comp(const Node & a,const Node & b)
 78 {
 79     return    a.cost < b.cost;
 80 }
 81 
 82 int    kruskal(void)
 83 {
 84     int    count = 0,ans = 0;
 85     bool    flag = true;
 86 
 87     for(int i = 0;i < NUM;i ++)
 88         if(!same(G[i].from,G[i].to))
 89         {
 90             if(i + 1 < NUM && G[i].cost == G[i + 1].cost && (G[i].from == G[i + 1].from || G[i].from == G[i + 1].to || 
 91                G[i].to == G[i + 1].to || G[i].to == G[i + 1].from) && !same(G[i + 1].from,G[i + 1].to))
 92             {
 93                 flag = false;
 94                 break;
 95             }
 96             unite(G[i].from,G[i].to);
 97             count ++;
 98             ans += G[i].cost;
 99             if(count == N - 1)
100                 break;
101         }
102     if(flag)
103         printf("%d
",ans);
104     else
105         puts("Not Unique!");
106 }