POJ 1679 The Unique MST

独特最小生成树与图论问题解决

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原文链接：http://www.cnblogs.com/372465774y/archive/2012/08/27/2658098.html

本文探讨了在给定连通无向图的情况下，如何判断其唯一最小生成树的存在性。通过实例和算法解析，展示了如何利用克鲁斯卡尔算法来寻找最小生成树，并通过比较不同路径的成本来确定是否唯一。

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15153		Accepted: 5241

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!

Source

POJ Monthly--2004.06.27 srbga@POJ

//这么久没刷题了、回到学校了、第一次写次小生成树，开心的是1A，不过写代码速度下降了
//I am back
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define N 102
using namespace std;
struct node
{
    int a,b,w;
    bool visit;
    bool operator <(const node &t)const
    {
        return w<t.w;
    }
};
struct Link
{
    int to;
    int next;
};
Link link[N];
int len[N][N];
int fa[N];
node Ed[5055];
int n,m,Min;
int Find(int x)
{
    if(x!=fa[x])
    {
      return fa[x]=Find(fa[x]);
    }
    return x;
}

bool merge(int x,int y,int i)
{
    x=Find(x);
    y=Find(y);
    if(x!=y)
    {
        int u,v;
        for(u=link[x].to;;u=link[u].next)
          {
            for(v=link[y].to;;v=link[v].next)
              {
                  len[u][v]=len[v][u]=Ed[i].w;
                  if(link[v].next==-1) break;
              }
            if(link[u].next==-1) break;
          }
       Ed[i].visit=1;
       link[u].next=y;
       fa[y]=x;
      return true;
    }
    return false;
}
void Kruskal()
{
    int i;
    for(Min=i=0;i<m;i++)
      if(merge(Ed[i].a,Ed[i].b,i))
        Min+=Ed[i].w;
}
int main()
{
    int t,i;
    scanf("%d",&t);
    while(t--)
    {
       scanf("%d %d",&n,&m);
       for(i=1;i<=n;i++)
        fa[i]=i,link[i].to=i,link[i].next=-1;
       for(i=0;i<m;i++)
       {
           scanf("%d %d %d",&Ed[i].a,&Ed[i].b,&Ed[i].w);
           Ed[i].visit=0;
       }
       sort(Ed,Ed+m);
       Kruskal();
       int Mi=-1,tp;
       for(i=0;i<m;i++)
        if(!Ed[i].visit)
        { tp=Min+Ed[i].w-len[Ed[i].a][Ed[i].b];
            if(tp>Min)
             Mi=tp;
            else
            {
                Mi=tp;break;
            }
        }
        if(Min==Mi)
         printf("Not Unique!\n");
        else
         printf("%d\n",Min);

    }
    return 0;
}

转载于:https://www.cnblogs.com/372465774y/archive/2012/08/27/2658098.html