POJ 1679 The Unique MST（次小生成樹）

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16984		Accepted: 5892

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

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

POJ Monthly--2004.06.27 srbga@POJ

 

 

 

判斷最小生成樹是不是唯一。

 

可以求次小生成樹，如果相等說明不唯一

 

//============================================================================
// Name        : POJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

/*
 * 次小生成樹
 * 求最小生成樹時，用數組Max[i][j]來表示MST中i到j最大邊權
 * 求完后，直接枚舉所有不在MST中的邊，替換掉最大邊權的邊，更新答案
 * 點的編號從0開始
 */
const int MAXN=110;
const int INF=0x3f3f3f3f;
bool vis[MAXN];
int lowc[MAXN];
int pre[MAXN];
int Max[MAXN][MAXN];//Max[i][j]表示在最小生成樹中從i到j的路徑中的最大邊權
bool used[MAXN][MAXN];
int Prim(int cost[][MAXN],int n)
{
    int ans=0;
    memset(vis,false,sizeof(vis));
    memset(Max,0,sizeof(Max));
    memset(used,false,sizeof(used));
    vis[0]=true;
    pre[0]=-1;
    for(int i=1;i<n;i++)
    {
        lowc[i]=cost[0][i];
        pre[i]=0;
    }
    lowc[0]=0;
    for(int i=1;i<n;i++)
    {
        int minc=INF;
        int p=-1;
        for(int j=0;j<n;j++)
            if(!vis[j]&&minc>lowc[j])
            {
                minc=lowc[j];
                p=j;
            }
        if(minc==INF)return -1;
        ans+=minc;
        vis[p]=true;
        used[p][pre[p]]=used[pre[p]][p]=true;
        for(int j=0;j<n;j++)
        {
            if(vis[j])Max[j][p]=Max[p][j]=max(Max[j][pre[p]],lowc[p]);
            if(!vis[j]&&lowc[j]>cost[p][j])
            {
                lowc[j]=cost[p][j];
                pre[j]=p;
            }
        }
    }
    return ans;
}
int ans;
int smst(int cost[][MAXN],int n)
{
    int Min=INF;
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++)
            if(cost[i][j]!=INF && !used[i][j])
            {
                Min=min(Min,ans+cost[i][j]-Max[i][j]);
            }
    if(Min==INF)return -1;//不存在
    return Min;
}
int cost[MAXN][MAXN];
int main()
{
    int T;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        int u,v,w;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                if(i==j)cost[i][j]=0;
                else cost[i][j]=INF;
            }
        while(m--)
        {
            scanf("%d%d%d",&u,&v,&w);
            u--;v--;
            cost[u][v]=cost[v][u]=w;
        }
        ans=Prim(cost,n);
        if(ans==-1)
        {
            printf("Not Unique!\n");
            continue;
        }
        if(ans==smst(cost,n))printf("Not Unique!\n");
        else printf("%d\n",ans);
    }
    return 0;
}