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
Problem 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'.
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!
Source
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();
}
}
