poj 1679 The Unique MST

最新推荐文章于 2020-08-14 19:39:00 发布

原创最新推荐文章于 2020-08-14 19:39:00 发布 · 426 阅读

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#tree #graph #properties #each #integer #input

本文介绍了一种判断给定图的最小生成树是否唯一的算法思路。通过使用Kruskal算法求解最小生成树，并通过删除已选边来验证其唯一性。特别地，文章还提出了一种优化方案，仅检查权重相同的边。

http://acm.pku.edu.cn/JudgeOnline/problem?id=1679

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

Sample Output

3
Not Unique!

看到网上大家都是说求次小生成树，对这个也不太熟，我的想法是利用MST的性质，边数相同，MST边中有相等边的情况，简单实现是先用Prim或者

kruskal算法求出一组解，并记录所选择的边，为验证唯一直接采用删边法，逐条删边，出现相同的解则退出，得出唯一的结论则要访问到最后一条边

不过有个小优化是 只探查存在相同权值的边，这样开销应该小很多，具体实现因为时间的关系 现在就省了  留下某牛牛代码以后参考：

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef struct
{
int x;
int y;
int w;
}edge;
edge s[10010];
int top,t[10010];
bool cmp(edge s1,edge s2)
{
return s1.w<s2.w;
}
int kruscal(int n,int m,int x)
{
int i,j,a,b,tag[110],tem,sum,k;
for(i=0;i<n;i++)
{
   tag[i]=i;
}
k=1;j=0;sum=0;
while(k<n)
{
   a=s[j].x-1;
   b=s[j].y-1;
   if(j==x)
   {
    j++;
    continue;
   }
   if(tag[a]!=tag[b])
   {
    if(x==-1)
    {
     t[top]=j;
     top++;
    }
    tem=tag[b];
    k++;sum+=s[j].w;
    for(i=0;i<n;i++)
    {
     if(tag[i]==tem)
     {
      tag[i]=tag[a];
     }
    }
   }
   j++;
}
return sum;
}  
int main()
{
freopen("in.txt","r",stdin);
int p,n,m,l,cmin,min,i,key;
while(scanf("%d",&p)!=EOF)
{
   while(p--)
   {
    scanf("%d%d",&n,&m);
    for(i=0;i<m;i++)
    {
     scanf("%d%d%d",&s[i].x,&s[i].y,&s[i].w);
    }
    sort(s,s+m,cmp);
    /*for(i=0;i<m;i++)
    {
     printf(" %d ",s[i].w);
    }
    printf("/n");*/
    top=0;
    min=kruscal(n,m,-1);
    key=top;
    for(l=0;l<key;l++)
    {
     cmin=kruscal(n,m,t[l]);
     if(cmin==min)
     {
      printf("Not Unique!/n");
      break;
     }
    }
    if(l==top&&cmin!=min)
    {
     printf("%d/n",min);
    }
   }
}
}