POJ 1679 (次最小生成树)

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概述

题目：

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

Sample Output

3
Not Unique!

思路：

先用克鲁斯卡尔算法求出其最小生成树，并记录在最小生成树上的每一条边。还要记录最小生成树的长度。

然后去掉最小生成树上的一条边，遍历其上所有边，看是否还存在长度为最小生成树长度的树。若存在，则不唯一。

若不存在，则输出最小生成树的长度。

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#define Max 99999999
using namespace std;
int t,n,m;
int a[102],st[102];
struct node{
int x;
int y;
int d;
}g[10000],gg[10000];
void init()
{
int i,j;
for(i=0;i<=n;i++)
{
a[i]=i;
}
}
int find(int x)
{
if(x==a[x])return a[x];
a[x]=find(a[x]);
return a[x];
}
bool Union(int x,int y)
{
int xx=find(x);
int yy=find(y);
if(xx==yy)return false;
a[xx]=yy;
return true;
}
bool cmp(node p,node q)
{
return p.d <q.d ;
}
int main()
{
cin>>t;
while(t--)
{
scanf("%d%d",&n,&m);
int i,j,k,l;
init();
for(i=0;i<m;i++)
{
scanf("%d%d%d",&j,&k,&l);
g[i].x =j;g[i].y =k;
g[i].d =l;
}
int sum;
sort(g,g+m,cmp);
for(i=0,k=0,sum=0;i<m;i++)
{
if(Union(g[i].x ,g[i].y ))
{
sum=sum+g[i].d ;
st[k]=i;k++;//用st数组记录边
}
}
int ans;
for(i=0;i<k;i++)//遍历去掉一条边
{
ans=0;
l=g[st[i]].d ;
g[st[i]].d =Max;
for(j=0;j<m;j++)
{
gg[j].x =g[j].x ;
gg[j].y =g[j].y ;
gg[j].d =g[j].d ;
}
sort(gg,gg+m,cmp);
init();
for(j=0;j<m;j++)
{
if(Union(gg[j].x ,gg[j].y ))
{
ans=ans+gg[j].d ;
}
}
g[st[i]].d =l;
if(ans==sum)break;
}
if(i<k)
printf("Not Unique!n");
else printf("%dn",sum);
}
return 0;
}