POJ1679The Unique MST（次小生成树）

The Unique MST
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 25203   Accepted: 8995

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!

题意：问一棵最小生成树是否唯一
找次小生成树，如果相等不唯一，否则唯一

次小生成树：就是最小生成树换一条边而成的生成树；用maxd【x】【y】存储最小生成树两个节点(x,y)路径中最大的那条边的权值,也就是最小生成树中x-z-...-y中那个最大的那一个权值，然后就可以换边了，到底换哪一个边就从不在生成树中的边一个一个枚举咯，假设x-y，如果要用x-y替换x-z...-y肯定得替换x-z...-y中权值最大的那条边才能让最后得到结果只比x-z...-y小一点，也就是仅次于
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <cstring>
  5 using namespace std;
  6 const int MAX = 110;
  7 const int INF = 100000000;
  8 int n,m,ans,cnt;
  9 int edge[MAX][MAX],vis[MAX],used[MAX][MAX],dist[MAX];
 10 int pre[MAX],maxd[MAX][MAX];
 11 void prime()
 12 {
 13     memset(vis,0,sizeof(vis));
 14     memset(used,false,sizeof(used));
 15     memset(maxd,0,sizeof(maxd));
 16     for(int i = 1; i <= n; i++)
 17     {
 18         dist[i] = edge[1][i];
 19         pre[i] = 1;
 20     }
 21     dist[1] = 0;
 22     vis[1] = 1;
 23     pre[1] = -1;
 24     for(int i = 1; i < n; i++)
 25     {
 26         int minn = INF,pos = -1;
 27         for(int j = 1; j <= n; j++)
 28         {
 29             if(vis[j] == 0 && dist[j] < minn)
 30             {
 31                 minn = dist[j];
 32                 pos = j;
 33             }
 34         }
 35         if(minn == INF)
 36         {
 37             ans = INF;
 38             return;
 39         }
 40         ans += minn;
 41         vis[pos] = 1;
 42         used[pos][pre[pos]] = used[ pre[pos] ][pos] = true;
 43         for(int j = 1; j <= n; j++)
 44         {
 45             if(vis[j])
 46                 maxd[j][pos] = maxd[pos][j] = max(dist[pos], maxd[j][pre[pos]]);
 47             if(vis[j] == 0 && dist[j] > edge[pos][j])
 48             {
 49                 dist[j] = edge[pos][j];
 50                 pre[j] = pos;
 51             }
 52         }
 53     }
 54 }
 55 void smst()
 56 {
 57     cnt = INF;
 58     for(int i = 1; i <= n; i++)
 59     {
 60         for(int j = i + 1; j <= n; j++)
 61         {
 62             if(used[i][j] == false && edge[i][j] != INF)
 63             {
 64                 cnt = min(cnt, ans - maxd[i][j] + edge[i][j]);
 65             }
 66         }
 67     }
 68 }
 69 int main()
 70 {
 71     int t;
 72     scanf("%d", &t);
 73     while(t--)
 74     {
 75         for(int i = 1; i <= MAX; i++)
 76             for(int j = 0; j <= MAX; j++)   //初始化的时候把MAX，写成了n, orz....
 77             {
 78                 if(i != j)
 79                     edge[i][j] = INF;
 80                 else
 81                     edge[i][i] = 0;
 82             }
 83         scanf("%d%d", &n,&m);
 84         for(int i = 0; i < m; i++)
 85         {
 86             int x,y,w;
 87             scanf("%d%d%d", &x, &y, &w);
 88             edge[x][y] = edge[y][x] = w;
 89         }
 90         ans = 0;
 91         prime();
 92         smst();
 93         if(ans == INF)
 94         {
 95             printf("Not Unique!\n");
 96             continue;
 97         }
 98         if(cnt == ans)
 99         {
100             printf("Not Unique!\n");
101         }
102         else
103         {
104             printf("%d\n",ans);
105         }
106     }
107     return 0;
108 }
View Code

 

 
posted @ 2015-11-22 17:00  zhaop  阅读(181)  评论(0)  编辑  收藏  举报