POJ 1679 The Unique MST 确定MST是否唯一

　　　　　　　　　　　　　　　　　　　　The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22817		Accepted: 8114

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

 

 

 

题意：给出一个图，若图中的MST图只有一个，输入最小代价，若不唯一，输入Not Unique!

 

 

O(n^3)

 

先求出一个MST图，最小代价ans，并且记录MST图的边，然后枚举每一条MST图的边

若去掉该条边后，该图的最小代价还是ans，说明MST图不止一个，则break

最后若枚举每一条边后，新的最小代价总是大于ans

说明该图的MST图只有一个。

 

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int MAXN=103;
const int INF=0x3f3f3f3f;

int cost[MAXN][MAXN];
int pre[MAXN];
int lowc[MAXN];
bool vis[MAXN];

void init(int N)
{
    for(int i=1;i<=N;i++)
    {
        for(int j=1;j<=N;j++)
        {
            if(i==j)
                cost[i][j]=0;
            else
                cost[i][j]=INF;
        }
    }
}

int prim(int N,int cnt)
{
    memset(vis,false,sizeof(vis));
    if(cnt)
        memset(pre,-1,sizeof(pre));
    vis[1]=true;
    int ans=0;
    for(int i=1;i<=N;i++)
    {
        lowc[i]=cost[1][i];
        if(cnt)
            pre[i]=1;
    }

    for(int j=1;j<N;j++)
    {
        int p=-1;
        int minc=INF;
        for(int i=1;i<=N;i++)
        {
            if(!vis[i]&&lowc[i]<minc)
            {
                minc=lowc[i];
                p=i;
            }
        }
        if(p==-1)
            return -1;
        ans+=minc;
        vis[p]=true;
        for(int i=1;i<=N;i++)
        {
            if(!vis[i]&&cost[p][i]<lowc[i])
            {
                lowc[i]=cost[p][i];
                if(cnt)
                    pre[i]=p;
            }
        }
    }
    return ans;
}

int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {
        int N,M;
        scanf("%d%d",&N,&M);
        init(N);

        for(int i=0;i<M;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            cost[u][v]=cost[v][u]=w;
        }

        int ans=prim(N,1);

        for(int i=2;i<=N;i++)
        {
            int v=pre[i];
            int tmp=cost[i][v];
            cost[i][v]=cost[v][i]=INF;
            int sum=prim(N,0);
            if(ans==sum)
            {
                ans=-1;
                break;
            }
            cost[i][v]=cost[v][i]=tmp;
        }
        if(ans==-1)
            printf("Not Unique!
");
        else
            printf("%d
",ans);
    }
    return 0;
}