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!
裸的判定最小生成樹是否唯一
做法
1，對圖中每一條邊，如果存在與之相等的其他的邊，則標(biāo)記這條邊
2，求一次最小生成樹，得到weight1，作為比較用
3，對于最小生成樹中的每一條邊，檢查這條邊有沒有與之相同的，如果有，則刪掉這條邊，
再求最小生成樹，如果相等，則說明最小生成樹不唯一
判斷完所有要判斷的邊后任然不相等，說明最小生成樹唯一 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所在樹節(jié)點數(shù)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