POJ1679-The Unique MST

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本文介绍了一种算法，用于判断给定图的最小生成树是否唯一。通过使用Kruskal算法两次，分别找到最小生成树及其总权重，并检查移除每条边后的最小生成树总权重是否相同来实现。

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The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 29566		Accepted: 10583

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'.

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!

Source

POJ Monthly--2004.06.27 srbga@POJ

题意：给你N个点M条边的图，图的最小生成树是否唯一

解题思路：次小生成树

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

#define MAXV 110
#define MAXE 10100

struct node
{
    int s,e,w,flag;
}x[MAXE];

int n,m;
int f[MAXV];

int cmp(node a,node b)
{
    return a.w<b.w;
}

int Find(int x)
{
    if(f[x]==x) return x;
    else return f[x]=Find(f[x]);
}

bool Union(int a,int b)
{
    int aa=Find(a);
    int bb=Find(b);
    if(aa==bb) return 0;
    f[aa]=bb;
    return 1;
}

int kruskal()
{
    int sum=0,cnt=0;
    for(int i=0; i<=n; i++) f[i]=i;
    for(int i=0; i<m; i++)
    {
        if(Union(x[i].s,x[i].e))
        {
            sum+=x[i].w;
            x[i].flag=1;
            cnt++;
            if(cnt==n-1) break;
        }
    }
    if(cnt==n-1) return sum;
    return -1;
}

int kruskalt()
{
    int sum=0,cnt=0;
    for(int i=0;i<=n;i++) f[i]=i;
    for(int i=0;i<m;i++)
    {
        if(Union(x[i].s,x[i].e))
        {
            sum+=x[i].w;
            cnt++;
            if(cnt==n-1) break;
        }
    }
    if(cnt==n-1) return sum;
    return -1;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&x[i].s,&x[i].e,&x[i].w);
            x[i].flag=0;
        }
        sort(x,x+m,cmp);
        int ans=kruskal();
        int flag=0;
        for(int i=0; i<m; i++)
        {
            if(x[i].flag)
            {
                int s=x[i].s;
                x[i].s=x[i].e;
                int tmp=kruskalt();
                if(tmp!=-1&&ans==tmp)
                {
                    flag=1;
                    break;
                }
                x[i].s=s;
            }
        }
        if(!flag) printf("%d\n",ans);
        else printf("Not Unique!\n");
    }
    return 0;
}