POJ1679 The Unique MST[次小生成树]

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最新推荐文章于 2019-05-04 17:33:50 发布

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本文介绍了一种判断给定无向图是否具有唯一最小生成树(MST)的方法，并提供了具体的算法实现。通过Kruskal算法构造图并进行深度优先搜索(DFS)，递归计算各节点间的最大边权重，以此来判断次小生成树是否与最小生成树相同。

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The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 28673		Accepted: 10239

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

求次小生成树，看看是不是和最小一样

方法：

kruskal求MST同时建图，dfs转有根树同时递推f[i][j]为i到j在树上的路径中权值最大是多少O(n^2)

次小一定是最小加一条边减一条边得到，枚举加哪一条边比较w和f[u][v]

//
//  main.cpp
//  poj1679
//
//  Created by Candy on 10/11/2016.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=105,M=10005,INF=1e9;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,m,u,v,w;
struct edge{
    int v,w,ne;
}e[N<<1];
int h[N],cnt=0;
inline void ins(int u,int v,int w){
    cnt++;
    e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt;
    cnt++;
    e[cnt].v=u;e[cnt].w=w;e[cnt].ne=h[v];h[v]=cnt;
}
struct data{
    int u,v,w;
    bool operator <(const data &r)const{return w<r.w;}
}a[M];
int fa[N];
inline int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int use[N];
int kruskal(){
    sort(a+1,a+1+m);
    int cnt=0,ans=0;
    for(int i=1;i<=m;i++){
        int u=a[i].u,v=a[i].v,w=a[i].w;
        int f1=find(u),f2=find(v);
        if(f1==f2) continue;
        fa[f1]=f2;
        ins(u,v,w);
        use[i]=1;
        cnt++; ans+=w;
        if(cnt==n-1) break;
    }
    return ans;
}
int f[N][N],vis[N];
void dfs(int u){
    vis[u]=1;
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v,w=e[i].w;
        if(vis[v]) continue;
        for(int x=1;x<=n;x++) if(vis[x]) f[x][v]=f[v][x]=max(f[x][u],w);
        dfs(v);
    }
}
void sol(){
    cnt=0;memset(h,0,sizeof(h));
    memset(f,0,sizeof(f));
    memset(vis,0,sizeof(vis));
    memset(use,0,sizeof(use));
    for(int i=1;i<=n;i++) fa[i]=i;
    
    int ans=kruskal();
    dfs(1);
    int mn=INF;
    for(int i=1;i<=m;i++) if(!use[i]){
        int u=a[i].u,v=a[i].v,w=a[i].w;//printf("hi %d %d %d  %d\n",u,v,w,f[u][v]);
        mn=min(mn,w-f[u][v]);
    }
    if(mn==0) puts("Not Unique!");
    else printf("%d\n",ans);
}
int main(int argc, const char * argv[]){
    int T=read();
    while(T--){
        n=read();m=read();
        for(int i=1;i<=m;i++){a[i].u=read();a[i].v=read();a[i].w=read();}
        sol();
    }
    return 0;
}