The Unique MST(最小生成树的唯一性判断)

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!

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+5;
typedef long long ll;
using namespace std;
int pre[maxn],n,m,first;
 
struct node
{
    int x,y,val;
    int u;
    int e;
    int d;
} p[maxn];
int find(int x)
{
    if(pre[x]==x)
    {
        return x;
    }
    else
    {
        return pre[x]=find(pre[x]);
    }
}
int prime()
{
    int i,j,k,sum,num;
    sum=0;num=0;
    for(i=1;i<=n;i++) 
    pre[i]=i;
    for(i=1;i<=m;i++) {
        if(p[i].d) continue;
        int fx=find(p[i].x);
        int fy=find(p[i].y);
        if(fx!=fy) {
            num++;
            pre[fx]=fy;
            sum+=p[i].val;
            if(first) 
            p[i].u=1;
        }
        if(num==n-1) break;
    }
    return sum;
}
bool cmp(node x,node y)
{
    if(x.val<y.val)
    return true;
    else
    return false;
}
int main()
{
    int k,u,v,w,sum1,sum2;
    int T;
    scanf("%d",&T);
    while(T--) 
    {
        sum1=sum2=0;
        memset(p,0,sizeof(p));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++) 
        {
            scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].val);
        }
        for(int i=1;i<=m;i++) {
            for(int j=i+1;j<=m;j++) 
            {
                if(p[i].val==p[j].val) p[i].e=1;
            }
        }
        sort(p+1,p+1+m,cmp);
        first=1;
        sum1=prime();
        first=0;
        bool flag=false;
        for(int i=1;i<=m;i++) 
        {
            if(p[i].u && p[i].e) 
            {
                
                p[i].d=1;
                sum2=prime();
                if(sum1==sum2) 
                {
                    flag=true;
                    printf("Not Unique!\n");
                    break;
                }
            }
        }
        if(!flag) 
        printf("%d\n",sum1);
    }
}

来源：https://www.cnblogs.com/Staceyacm/p/11135078.html

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最小生成树