POJ_1679_The Unique MST(次小生成树)

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!

题意：问最小生成树是否唯一。

分析：求次小生成树，推断次小生成树和最小生成树是否相等。

求次小生成树的步骤：

(1)先用Prime求出最小生成树MST，在Prime的同一时候用一个矩阵mmax[ ][ ]记录在MST中连接随意两点u,v的唯一路径中权

值最大的那条边的权值。做法：Prime是每次添加一个节点t。用该点新加入MST的边与它前一个加入MST的点的mmax的值做比较。

(2)枚举最小生成树以外的边，并删除该边所在环上权值最大的边。

(3)取得的全部生成树中权值最小的一棵即为所求。

算法的时间复杂度为O(n^2)。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define maxn 111
#define inf 0x3f3f3f3f

int map[maxn][maxn],mmax[maxn][maxn];//map邻接矩阵存图,mmax示最小生成树中i到j的最大边权 
bool used[maxn][maxn];//判断该边是否加入最小生成树
int pre[maxn],dis[maxn];//pre用于mmax的构建，装前一个放入MST的结点，dis用于构建MST 

void init(int n)
{
    for (int i=1;i<=n;i++)//图初始化 
    {
        for (int j=1;j<=n;j++)
        {
            if (i==j)
            {
                map[i][j]=0;
            }
            else
            {
                map[i][j]=inf;
            }
        }
    }
}

void read(int m)
{
    int u,v,w;
    for (int i=0;i<m;i++)//读入图 
    {
        scanf("%d%d%d",&u,&v,&w);
        map[u][v]=map[v][u]=w;
    }
}
int prime(int n)//构建MST 
{
    int ans=0;
    bool vis[maxn];
    memset(vis,false,sizeof(vis));
    memset(used,false,sizeof(used));
    memset(mmax,0,sizeof(mmax));
    for (int i=2;i<=n;i++)
    {
        dis[i]=map[1][i];
        pre[i]=1;//1点为第一个放入MST的点，先设为所有点的前驱结点 
    }
    pre[1]=0;
    dis[1]=0;
    vis[1]=true;
    for (int i=2;i<=n;i++)
    {
        int min_dis=inf,k;
        for (int j=1;j<=n;j++)
        {
            if (vis[j]==0&&min_dis>dis[j]) 
            {
                min_dis=dis[j];
                k=j;
            }
        }
        if (min_dis==inf)//如果不存在最小生成树
        {
            return -1;
        }
        ans+=min_dis;
        vis[k]=true;
        used[k][pre[k]]=used[pre[k]][k]=true;//标记为放入MST的点 
        for (int j=1;j<=n;j++)
        {
            if (vis[j])
            {
                mmax[j][k]=mmax[k][j]=max(mmax[j][pre[k]],dis[k]);//最小生成树环的最大边
            }
            if (!vis[j]&&dis[j]>map[k][j])
                {
                    dis[j]=map[k][j];
                    pre[j]=k;
                }
        }
    }
    return ans;//最小生成树的权值之和
}
int smst(int n,int min_ans)//min_ans 是最小生成树的权值和
{
    int ans=inf;
    for (int i=1;i<=n;i++)//枚举最小生成树之外的边 
    {
        for (int j=i+1;j<=n;j++)
        {
            if (map[i][j]!=inf&&!used[i][j])
            {
                ans=min(ans,min_ans+map[i][j]-mmax[i][j]);//该边次小MST的权值为MST加上该边再减去该边所在环的最大MST边 
            }
        }
    }
    if (ans==inf)
    {
        return -1;
    }
    return ans;
}
void solve(int n)
{
    int ans=prime(n);
    if (ans==-1)
    {
        puts("Not Unique!");
        return;
    }
    if (smst(n,ans)==ans)//次小MST权值等于MST说明MST不唯一 
    {
        printf("Not Unique!
");
    }
    else 
    {
        printf("%d
",ans);
    }
}
int main()
{
    int t,n,m;
    
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&m);
        init(n);
        read(m);
        solve(n);
    }
    
    return 0;
}