poj 1679 次小生成树

http://poj.org/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

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!

/**
poj 1679 次小生成树
题目大意：给定一个无向图，问它的最小生成树是否唯一
解题思路：利用prim算法求出最小生成树，并且用mlen数组标记在已知的生成树上i和j两点之间的经过边的最大权值，最后枚举所有没有在最小生成树上的
          边，加上该边，并且减去对应mlen，得到的结果就是必须用当前边的最小生成树，所有之中最小的就是次小生成树的大小，和最小生成树进行比较即可
*/
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int maxn=1010;

struct note
{
    int x,y,z;
}p[maxn];

int a[maxn][maxn],dis[maxn];
int pre[maxn],n,m;

int flag[maxn][maxn],vis[maxn];
int  mlen[maxn][maxn];

int prim(int u)
{
    int sum=0;
    memset(flag,0,sizeof(flag));
    memset(vis,0,sizeof(vis));
    memset(mlen,0,sizeof(mlen));
    for(int i=1; i<=n; i++)
    {
        dis[i]=a[u][i];
        pre[i]=u;
    }
    vis[u]=1;
    for(int i=1; i<n; i++)
    {
        int minn=0x3f3f3f3f;
        int v=-1;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j]&&dis[j]<minn)
            {
                v=j;
                minn=dis[j];
            }
        }
        if(v!=-1)
        {
            sum+=dis[v];
            flag[v][pre[v]]=flag[pre[v]][v]=1;
            vis[v]=1;
            for(int k=1; k<=n; k++)
            {
                if(vis[k]&&k!=v)
                {
                    mlen[v][k]=mlen[k][v]=max(mlen[k][pre[v]],dis[v]);
                }
                if(!vis[k]&&a[v][k]<dis[k])
                {
                    dis[k]=a[v][k];
                    pre[k]=v;
                }
            }
        }
    }
    return sum;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j)
                    a[i][j]=0;
                else
                    a[i][j]=0x3f3f3f3f;
            }
        }
        for(int i=1; i<=m; i++)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            a[x][y]=a[y][x]=w;
        }
        int ans=prim(1);
        int cnt=0x3f3f3f3f;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(a[i][j]!=0x3f3f3f3f&&!flag[i][j])
                {
                    cnt=min(cnt,ans-mlen[i][j]+a[i][j]);
                }
            }
        }
        if(cnt!=ans)
            printf("%d\n",ans);
        else
            puts("Not Unique!");
    }
    return 0;
}