hdu 1679 The Unique MST （克鲁斯卡尔）
The Unique MST
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24152 | Accepted: 8587 |
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!
题目链接：http://poj.org/problem?
id=1679
题目大意：n个点m条路。给出每条路以及边权。推断最小生成树是否是唯一的。
解题思路：克鲁斯卡尔，推断是否存在等效边。这题数据太弱了。我推断等效边的方法不太对，竟然过了= =
代码例如以下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int fa[102];
struct EG
{
int u,v,w;
}eg[5005];
void get_fa()
{
for(int i=0;i<105;i++)
fa[i]=i;
}
int find (int x)
{
return x==fa[x]?x:fa[x]=find(fa[x]);
}
void Union(int a,int b)
{
int a1=find(a);
int b1=find(b);
if(a1!=b1)
fa[a1]=b1;
}
int cmp(EG a,EG b)
{
return a.w<b.w;
}
int main(void)
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m,ans=0,p=0,cnt=0;
get_fa();
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&eg[i].u,&eg[i].v,&eg[i].w);
}
sort(eg,eg+m,cmp);
for(int i=0;i<m;i++)
{
if(find(eg[i].u)!=find(eg[i].v))//假设当前边须要增加且下一条边也须要增加且它们权值相等即为等效边
{
if(i+1<m&&find(eg[i+1].u)!=find(eg[i+1].v)&&eg[i].w==eg[i+1].w)
{
p=1;
break;
}
Union(eg[i].u,eg[i].v);
ans+=eg[i].w;
cnt++;
}
if(cnt>=n)
break;
}
if(!p)
printf("%d\n",ans );
else
printf("Not Unique!\n");
}
}
