（poj）1679 The Unique MST 求最小生成树是否唯一 （求次小生成树与最小生成树是否一样）

Description

Given a connected undirected graph, tell if its minimum spanning tree is unique. 

Definition  (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:
. V' = V.
. T is connected and acyclic. 

Definition  (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 ( <= t <= ), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m ( <= n <= ), 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

Not Unique!

题意：问是否有一条唯一的最小生成树

方法：先求最小生成树，再求次小生成树，看是否相同，如相同则不唯一，否则唯一

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include <math.h>
#include<queue>
#define ll long long
#define INF 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof(a));
#define N 111
int Map[N][N],vis[N],used[N][N],f[N][N];
int s[N][N],es[N],dis[N];
int ans1,ans2,n;
using namespace std;
void init()
{
    for(int i=; i<=n; i++)
    {
        for(int j=; j<=n; j++)
            Map[i][j]=i==j?:INF;
    }
    met(vis,);///标记是否用过
    met(s,);///标记两点是否连通
    met(f,);
    met(es,-);///你找的该点的上一个点，就是父节点
    met(used,);///i，j这条边是最小生成树上的边
}
void prim()///查找最小生成树
{
    ans1=;
    for(int i=; i<=n; i++)
    {
        dis[i]=Map[][i];
        if(dis[i]!=INF)
            es[i]=;
    }
    vis[]=;
    for(int i=; i<n; i++)
    {
        int an=INF,k=-;
        for(int j=; j<=n; j++)
        {
            if(!vis[j] && an>dis[j])
                an=dis[k=j];
        }
        if(k==-)
            return ;
        ans1+=an;
        used[k][es[k]]=used[es[k]][k]=;
        for(int j=; j<=n; j++)
        {
            if(vis[j])
            {
                f[j][k]=max(f[j][es[k]],Map[es[k]][k]);
                f[k][j]=f[j][k];
            }

        }
        vis[k]=;
        for(int j=; j<=n; j++)
        {
            if(!vis[j] && dis[j]>Map[k][j])
            {
                dis[j]=Map[k][j];
                es[j]=k;
            }
        }
    }
}
void sond()///从最小生成树上减去一条边加上另一条较小边，生成次小生成树
{
    ans2=INF;
    for(int i=; i<=n; i++)
    {
        for(int j=; j<=n; j++)
        {
            if(s[i][j] && !used[i][j] && ans1+Map[i][j]-f[i][j]<ans2)
                ans2=ans1+Map[i][j]-f[i][j];
        }
    }
}

int main()
{
    int t,m,a,b,l;
    scanf("%d",&t);
    while(t--)
    {
        ans1=ans2=;
        scanf("%d %d",&n,&m);
        init();
        for(int i=; i<m; i++)
        {
            scanf("%d %d %d",&a,&b,&l);
            Map[a][b]=Map[b][a]=l;
            s[a][b]=s[b][a]=;
        }
        prim();
        sond();
        if(ans1==ans2)
            printf("Not Unique!\n");
        else
            printf("%d \n",ans1);
    }
    return ;
}