The Unique MST
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 13200 |
|
Accepted: 4575 |
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!
裸的判定最小生成树是否唯一
做法
1，对图中每一条边，如果存在与之相等的其他的边，则标记这条边
2，求一次最小生成树，得到weight1，作为比较用
3，对于最小生成树中的每一条边，检查这条边有没有与之相同的，如果有，则删掉这条边，
再求最小生成树，如果相等，则说明最小生成树不唯一
判断完所有要判断的边后任然不相等，说明最小生成树唯一 1
#include<algorithm>
2#include<cstdlib>
3using namespace std;
4#define maxn 101
5#define maxm 15000
6struct node
7
{
8
int u,v,w;
9 int equal,used,del;
10
} edge[maxm];
11int n,m;
12int parent[maxn];
13int first;
14void ufset()
15{
16 int i;
17 for(i=1; i<=n; i++) parent[i]=-1;
18}
19int find(int x)
20{
21 int s;
22 for(s=x; parent[s]>=0; s=parent[s]);
23 while(s!=x)
24
{
25 int tmp=parent[x];
26 parent[x]=s;
27 x=tmp;
28
}
29 return s;
30}
31void union1(int R1,int R2)
32{
33 int r1=find(R1),r2=find(R2);
34 int tmp=parent[r1]+parent[r2];
35 if (parent[r1]>parent[r2])//r2所在树节点数多于r1
36 {
37 parent[r1]=r2;
38 parent[r2]=tmp;
39 }
40 else
41 {
42 parent[r2]=r1;
43 parent[r1]=tmp;
44 }
45}
46int cmp(struct node a,struct node b)
47{
48 return a.w<b.w;
49}
50int kruskal()
51{
52 int sumweight=0,num=0;
53 int u,v;
54 ufset();
55 for(int i=0; i<m; i++)
56 {
57 if (edge[i].del==1)
58 {
59 continue;
60 }
61 u=edge[i].u;
62 v=edge[i].v;
63 if (find(u)!=find(v))
64 {
65 sumweight+=edge[i].w;
66 num++;
67 union1(u,v);
68 if (first)
69 {
70 edge[i].used=1;
71 }
72 }
73 if (num>=n-1)
74 {
75 break;
76 }
77 }
78 return sumweight;
79}
80int main()
81{
82 int t,i,j,k;
83 int u,v,w;
84 scanf("%d",&t);
85 for(i=1; i<=t; i++)
86 {
87 scanf("%d%d",&n,&m);
88 memset(edge,0,sizeof(edge));
89 for(j=0; j<m; j++)
90 {
91 scanf("%d%d%d",&u,&v,&w);
92 edge[j].u=u;
93 edge[j].v=v;
94 edge[j].w=w;
95 }
96 for(j=0; j<m; j++)
97 for(k=0; k<m; k++)
98 {
99 if (k==j) continue;
100 if (edge[j].w==edge[k].w) edge[j].equal=1;
101 }
102 sort(edge,edge+m,cmp);
103 first=1;
104 int weight1=kruskal(),weight2;
105 first=0;
106 for(j=0;j<m;j++)
107 {
108 if (edge[j].used==1&&edge[j].equal==1)
109 {
110 edge[j].del=1;
111 weight2=kruskal();
112 if (weight2==weight1)
113 {
114 printf("Not Unique!\n");
115 break;
116 }
117 edge[j].del=0;
118 }
119 }
120 if (j>=m)
121 {
122 printf("%d\n",weight1);
123 }
124 }
125 return 0;
126}
127