POJ - 1679_The Unique MST

The Unique MST

Time Limit: 1000MS Memory Limit: 10000K

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!

题意：判断最小生成树是不是唯一。
题解：首先判断能不能形成最小生成树，然后寻找次小生成树，判断是否等于最小生成树，如果不相等，则说明最小生成树唯一。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
#include <map>

using namespace std;

const int maxn = 105;
const int INF = 1e9+7;

int s[maxn][maxn],n,m;

int prim()
{
    int i,j,MIN,k,sum = 0;
    int dis[maxn],f[maxn],pre[maxn],maxx[maxn][maxn];/*maxx用来存储最小生成树的 i 到 j 的最大边权值*/
    int use[maxn][maxn];
    for(i=1;i<=n;i++)
    {
        dis[i] = s[1][i];
        pre[i] = 1;
        f[i] = 0;
    }
    memset(maxx,0,sizeof(maxx));
    memset(use,0,sizeof(use));
    f[1] = 1;
    for(i=1;i<n;i++)
    {
        MIN = INF;
        k = -1;
        for(j=1;j<=n;j++)
            if(!f[j]&&MIN>dis[j])
            {
                MIN = dis[j];
                k = j;
            }
        if(MIN==INF)
            return -1;
        //printf("MIN %d\n",MIN);
        f[k] = 1;
        sum += MIN;
        for(j=1;j<=n;j++)
        {
            if(f[j])
                maxx[k][j] = maxx[j][k] = max(maxx[pre[k]][j],dis[k]);/*（发现直接等于dis[k]也行）*/
            else
            {
                if(dis[j]>s[k][j])
                {
                    dis[j] = s[k][j];
                    pre[j] = k;
                }
            }
        }
    }
    //printf("%d\n",sum);
    MIN = INF;
    for(i=1;i<=n;i++)
    {
        for(j=i+1;j<=n;j++)
        {
            if(i!=pre[j]&&j!=pre[i]&&s[i][j]!=INF)
                MIN = min(MIN,sum-maxx[i][j]+s[i][j]);
        }
    }
    //printf("%d\n",MIN);
    if(MIN==sum)
        return -1;
    else
        return sum;
}

int main()
{
    int t,i,j,a,b,c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                s[i][j] = i==j?0:INF;
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(s[a][b]>c)
                s[a][b] = s[b][a] = c;
        }
        a = prim();
        if(a==-1)
            printf("Not Unique!\n");
        else
            printf("%d\n",a);
    }
    return 0;
}