POJ 1679 The Unique MST（最小生成树--prime||次小生成树）

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24586		Accepted: 8743

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!

判断最小生成树的唯一性

ac代码：
最小生成树解法：
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define MAXN 1010
#define INF 0xfffffff
#define max(a,b) a>b?a:b
#define min(a,b) a>b?b:a
using namespace std;
int pri[MAXN][MAXN];
int dis[MAXN];
int v[MAXN];
int sum,n,bz;
void prime()
{
	int i,k,M,j,q;
	memset(v,0,sizeof(v));
	for(i=1;i<=n;i++)
	dis[i]=pri[1][i];
	v[1]=1;
	sum=0;
	for(i=1;i<n;i++)
	{
		M=INF;
		q=0;
		for(j=1;j<=n;j++)
		{
			if(v[j]==0&&dis[j]<M)
			{
				M=dis[j];
				k=j;
			}
		}
		for(j=1;j<=n;j++)
		{
			if(v[j]&&pri[k][j]==M)
			{
				q++;
			}
		}
		v[k]=1;
		sum+=M;
		if(q>1)
		{
			bz=1;
		    break;
	    }
		if(M==INF)
		break;
		for(j=1;j<=n;j++)
		{
			if(v[j]==0&&dis[j]>pri[k][j])
			dis[j]=pri[k][j];
		}
	}
}
int main()
{
	int t,m,i,j,a,b,c;
	scanf("%d",&t);
	while(t--)
	{
		bz=0;
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			pri[i][j]=pri[j][i]=INF;
		}
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(pri[a][b]>c)
			pri[a][b]=pri[b][a]=c;
		}
		prime();
		if(bz)
		printf("Not Unique!\n");
		else
		printf("%d\n",sum);
	}
	return 0;
}
次小生成树解法：
#include<stdio.h>  
#include<string.h>  
#include<math.h>  
#include<iostream>  
#include<algorithm>  
#define INF 0xfffffff  
#define MAXN 550  
#define MAX(a,b) a>b?a:b  
#define MIN(a,b) a>b?b:a  
using namespace std;  
int v[MAXN];  
int pre[MAXN];  
int pri[MAXN][MAXN];    
int map[MAXN][MAXN];  
int connect[MAXN][MAXN];
int dis[MAXN];  
int n; 
int sum; 
void prime()  
{  
    int i,j,k,M;  
    memset(v,0,sizeof(v));  
    memset(map,0,sizeof(map)); 
    memset(connect,0,sizeof(connect)); 
    sum=0; 
    for(i=1;i<=n;i++)  
    {  
        dis[i]=pri[1][i];  
        pre[i]=1;  
    }  
    dis[1]=0;  
    v[1]=1;  
    for(i=1;i<n;i++)  
    {  
        M=INF;  
        for(j=1;j<=n;j++)  
        {  
            if(!v[j]&&dis[j]<M)  
            {  
                M=dis[j];  
                k=j;  
            }  
        }  
        if(M==INF)  
        break;  
        sum+=M;
        v[k]=1;   
        connect[k][pre[k]]=connect[pre[k]][k]=1; 
        for(j=1;j<=n;j++)  
        if(v[j]&&j!=k) 
            map[j][k]=map[k][j]=MAX(M,map[j][pre[k]]);  
        for(j=1;j<=n;j++)  
        {  
            if(!v[j]&&dis[j]>pri[k][j])  
            {  
                dis[j]=pri[k][j];  
                pre[j]=k;  
            }  
        }  
    }  
}  
int main()  
{  
    int t,m;  
    int i,j;  
    int a,b,c;  
    scanf("%d",&t);  
    while(t--)  
    {  
        int bz=0;  
        scanf("%d%d",&n,&m);  
        for(i=1;i<=n;i++)  
        for(j=1;j<=n;j++)  
        pri[i][j]=INF;    
        for(i=0;i<m;i++)  
        {  
            scanf("%d%d%d",&a,&b,&c);  
            pri[a][b]=pri[b][a]=c;  
        }  
        prime();  
        for(i=1;i<=n;i++)  
        {  
            for(j=i+1;j<=n;j++)  
            {   
                  if(i!=j&&!connect[i][j]&&pri[i][j]==map[i][j])
                  {
                      bz=1;
                      break;
                  }
            }  
        }  
        if(bz)  
        printf("Not Unique!\n");  
        else  
        printf("%d\n",sum);  
    }  
    return 0;  
}