poj-1679-The Unique MST

我不是女神ヾ 2022-05-26 11:43 88阅读 0赞

The Unique MST

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>34232</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;12463</td> 
  </tr> 
 </tbody> 
</table>

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][POJ Monthly--2004.06.27 srbga_POJ]

题目大意：判断最小生成树是否唯一

思路：用Kruskal 算出最小生成树的值，并记录每一条边，然后枚举去掉这些边 看其是否也能构成最小生成树且值相同。注意 在删边后，可能图构不成一棵树，得判断一下

代码：

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    struct ss
    {
        int u,v,w;
    } a[101010];
    int n,m,flag;
    int ans,fu[10101];
    void makeSet()
    {
        for(int i=1; i<=n; i++)
            fu[i]=i;
    }
    int findroot(int x)
    {
        if(x!=fu[x])
            fu[x]=findroot(fu[x]);
        return fu[x];
    }
    int cmp(ss a,ss b)
    {
        return a.w<b.w;
    }
    int hebing(int a,int b)
    {
        int x,y;
        x=findroot(a);
        y=findroot(b);
        if(x==y)
            return 0;
        fu[y]=x;
        return 1;
    }
    void Kruskal()
    {
        makeSet();
        sort(a,a+m,cmp);
        int path[101010],cnt=0;
        ans=0;
        for(int i=0; i<m; i++)
        {
            if(hebing(a[i].u,a[i].v))
            {
                path[cnt++]=i; //记录路径 
                ans+=a[i].w;
            }
        }
        for(int k=0; k<cnt; k++)//枚举去掉每一条边
        {
            makeSet();
            int sum=0,j=0;
            for(int i=0; i<m; i++)
            {
                if(i==path[k])
                    continue;
                if(hebing(a[i].u,a[i].v))
                {
                    sum+=a[i].w;
                    j++;
                }
            }
            if(j==n-1&&sum==ans)//判断是否能构成树 且 是否与最小生成树相等 
            {
                flag=0;
                return ;
            }
        }
    }
    
    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",&a[i].u,&a[i].v,&a[i].w);
            flag=1;
            Kruskal();
            if(flag)
                printf("%d\n",ans);
            else
                printf("Not Unique!\n");
        }
        return 0;
    }

[POJ Monthly--2004.06.27 srbga_POJ]: http://poj.org/searchproblem?field=source&key=POJ+Monthly--2004.06.27+srbga%40POJ