次小生成树（poj1679）

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20737		Accepted: 7281

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：判断最小生成树是否唯一；

分析：最小生成树的值和次小生成树的值相等则不唯一；

程序：

#include"stdio.h"
#include"string.h"
#define inf 100000000
#define M 1111
int G[M][M],maxd[M][M],use[M],dis[M],pre[M],vis[M][M];
int max(int a,int b)
{
    return a>b?a:b;
}
int min(int a,int b)
{
    return a<b?a:b;
}
int dij(int u,int n)
{
    int ans=0,i,j;
    memset(use,0,sizeof(use));
    memset(maxd,0,sizeof(maxd));//记录不在任意两点在在生成树的路径中的最长边
    memset(vis,0,sizeof(vis));//标记边是否在生成树里面
    for(i=1;i<=n;i++)
    {
        dis[i]=G[u][i];
        pre[i]=u;//记录父节点
    }
    dis[u]=0;
    use[u]=1;
    for(i=1;i<n;i++)
    {
        int mini=inf;
        int tep=-1;
        for(j=1;j<=n;j++)
        {
            if(!use[j]&&dis[j]<mini)
            {
                mini=dis[j];
                tep=j;
            }
        }
        if(tep==-1)break;
        use[tep]=1;
        vis[tep][pre[tep]]=vis[pre[tep]][tep]=1;
        ans+=mini;
        for(j=1;j<=n;j++)
        {
            if(!use[j]&&dis[j]>G[tep][j])
            {
                dis[j]=G[tep][j];
                pre[j]=tep;
            }
            if(j!=tep)
            maxd[tep][j]=maxd[j][tep]=max(mini,maxd[pre[tep]][j]);//更新
        }
    }
    return ans;
}
int main()
{
    int T,m,n,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            G[i][j]=inf;
        while(m--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            if(G[a][b]>c)
                G[a][b]=G[b][a]=c;
        }
        int ans=dij(1,n);
        int cnt=inf;
        for(i=1;i<=n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                if(G[i][j]<inf&&vis[i][j]==0)
                {
                    cnt=min(cnt,ans+G[i][j]-maxd[i][j]);
                }
            }
        }
        if(ans==cnt)
            printf("Not Unique!
");
        else
            printf("%d
",ans);
    }
    return 0;
}