poj1679The Unique MST 次小生成树

The Unique MST

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21580 Accepted: 7636

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

23 31 2 12 3 23 1 34 41 2 22 3 23 4 24 1 2

Sample Output

3Not Unique!

Source

POJ Monthly--2004.06.27 srbga@POJ

先求出最小生成树，记录最小生成树的每条边，除去这些边构造出的新的最小生成树总和最小的就是次小生成树了

code：

#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
#define Max 102
#define Maxx 5002


struct edge
{
    int u,v;
    int w;
}edges[Maxx];
int n,m,p[Max],sum;




bool cmp(edge a,edge b)
{
    if(a.w<b.w) return true;
    return false;
}


void build()
{
    for(int x=1;x<=n;x++)
        p[x]=x;
}


int look(int x)
{
    if(x!=p[x])
        p[x]=look(p[x]);
    return p[x];
}
bool uni;
void kruskal()
{
    int x,y;
    int a[Max],numm,k=0;
    build();
    sort(edges,edges+ m,cmp);
    sum=0;
    for(int i=0;i<m;i++)
    {
        x = look(edges[i].u);
        y = look(edges[i].v);
        if(x!=y)
        {
            p[y]=x;
            sum+=edges[i].w;
            a[k++]=i;
        }
    }
    numm=k;
    uni=true;
    int r,num;
    for(k=0;k<numm;k++)
    {
        build();
        r=0;
        num=0;
        for(int i=0;i<m;i++)
        {
            if(i==a[k]) continue;
            x=look(edges[i].u);
            y=look(edges[i].v);
            if(x!=y)
            {
                p[y]=x;
                r+=edges[i].w;
                num++;
            }
        }
        if(num!=numm) continue;
        if(r==sum)
        {
            uni=false;
            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", &edges[i].u, &edges[i].v, &edges[i].w);
        kruskal();
        if(!uni) printf("Not Unique!\n");
        else printf("%d\n",sum);
    }
    return 0;
}