The Unique MST--hdoj

                                              The Unique MST

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 25   Accepted Submission(s) : 8
[align=left]Problem Description[/align]
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'.

 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

3
Not Unique!

 

[align=left]Source[/align]
PKU

有时候最小生成树不止一个，举个例子，某一个点以min进入当前的集合，但是进去之后又发现他与好几个在集合中的点的距离都是min，这也就说明，他连哪一个点都是一样的，所以此时的最小生成树不止一个。代码：

#include<stdio.h>
#include<string.h>
#define INF 0xfffffff
int n,m;
int map[1010][1010],dis[1010],mark[1010];
void prim()
{
int i,min,flag,j,sum=0,k,w=1;
memset(mark,0,sizeof(mark));
mark[1]=1;
for(i=2;i<=n;i++)
{
min=INF;
flag=-1;
k=0;
for(j=1;j<=n;j++)
{
if(!mark[j]&&map[1][j]<min)
{
min=map[1][j];
flag=j;
}
}
for(j=1;j<=n;j++)
{
if(mark[j]&&map[flag][j]==min)
{/*因为mark[1]已经标记过了，所以此时如果出现两个及其以上的点，
到flog的距离都是min，当然，仅限于已经连过的点，此时最小生成树就不止一种*/
k++;
}
}
//printf("%d  ",k);
if(k>=1)
{
w=0;
break;
}
mark[flag]=1;
sum+=min;
for(j=1;j<=n;j++)
{
if(!mark[j]&&map[1][j]>map[flag][j])
map[1][j]=map[flag][j];
}
}
if(w)
printf("%d\n",sum);
else
printf("Not Unique!\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int i,j;
scanf("%d%d",&n,&m);
for(i=0;i<=n;i++)
{
for(j=0;j<=n;j++)
map[i][j]=INF;
//map[i][i]=0;
}
while(m--)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
map[a][b]=map[b][a]=c;
}
prim();
}
}