Slim Span

Given an undirected weighted graph G, you should find one of spanning trees specified as follows.

The graph G is an ordered pair (V, E), where V is a set of vertices {v1, v2, …, vn} and E is a set of undirected edges {e1, e2, …, em}. Each edge e ∈ E has its weight w(e).

A spanning tree T is a tree (a connected subgraph without cycles) which connects all the n vertices with n − 1 edges. The slimness of a spanning tree T is defined as the difference between the largest weight and the smallest weight among the n − 1 edges of T.

 
Figure 5: A graph G and the weights of the edges

For example, a graph G in Figure 5(a) has four vertices {v1, v2, v3, v4} and five undirected edges {e1, e2, e3, e4, e5}. The weights of the edges are w(e1) = 3, w(e2) = 5, w(e3) = 6, w(e4) = 6, w(e5) = 7 as shown in Figure 5(b).

Figure 6: Examples of the spanning trees of G

There are several spanning trees for G. Four of them are depicted in Figure 6(a)~(d). The spanning tree Ta in Figure 6(a) has three edges whose weights are 3, 6 and 7. The largest weight is 7 and the smallest weight is 3 so that the slimness of the tree Ta is 4. The slimnesses of spanning trees Tb, Tc and Td shown in Figure 6(b), (c) and (d) are 3, 2 and 1, respectively. You can easily see the slimness of any other spanning tree is greater than or equal to 1, thus the spanning tree Td in Figure 6(d) is one of the slimmest spanning trees whose slimness is 1.

Your job is to write a program that computes the smallest slimness.

Input

The input consists of multiple datasets, followed by a line containing two zeros separated by a space. Each dataset has the following format.

n	m
a1	b1	w1
	⋮
am	bm	wm

Every input item in a dataset is a non-negative integer. Items in a line are separated by a space. n is the number of the vertices and m the number of the edges. You can assume 2 ≤ n ≤ 100 and 0 ≤ m ≤ n(n − 1)/2. ak and bk (k = 1, …, m) are positive integers less than or equal to n, which represent the two vertices vak and vbkconnected by the kth edge ek. wk is a positive integer less than or equal to 10000, which indicates the weight of ek. You can assume that the graph G = (V, E) is simple, that is, there are no self-loops (that connect the same vertex) nor parallel edges (that are two or more edges whose both ends are the same two vertices).

Output

For each dataset, if the graph has spanning trees, the smallest slimness among them should be printed. Otherwise, −1 should be printed. An output should not contain extra characters.

Sample Input

4 5
1 2 3
1 3 5
1 4 6
2 4 6
3 4 7
4 6
1 2 10
1 3 100
1 4 90
2 3 20
2 4 80
3 4 40
2 1
1 2 1
3 0
3 1
1 2 1
3 3
1 2 2
2 3 5
1 3 6
5 10
1 2 110
1 3 120
1 4 130
1 5 120
2 3 110
2 4 120
2 5 130
3 4 120
3 5 110
4 5 120
5 10
1 2 9384
1 3 887
1 4 2778
1 5 6916
2 3 7794
2 4 8336
2 5 5387
3 4 493
3 5 6650
4 5 1422
5 8
1 2 1
2 3 100
3 4 100
4 5 100
1 5 50
2 5 50
3 5 50
4 1 150
0 0

Sample Output

1
20
0
-1
-1
1
0
1686
50

题意：给出n个点m条边，问能不能把这n个点构成一棵树，如果能输出树中最大边减最小边的最小值，如果不能输出-1.

思路：先利用结构体把这几条边从小到大排序，然后两层for循环尝试着构造树的枝，在尝试的过程中用并查集如果两点的根节点一样说明会成环，不能将这条枝加上，依次便利直到加满n-1条，那么最后加的一条减去第一天就是最大边减最小边，将循环走完不断更新这个值，最后输出即可。代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int m,n,flag,f[110];
struct node
{
    int a,b,l;
}list[10010];
bool cmp(node x,node y)
{
    return x.l<y.l;
}
void init()
{
    for(int i=1; i<=n; i++)
        f[i]=i;
}
int getf(int v)
{
    if(f[v]==v)
        return v;
    else
    {
        f[v]=getf(f[v]);
        return f[v];
    }
}
void merge(int u,int v)
{
    flag=0;
    int t1=getf(u);
    int t2=getf(v);
    if(t1!=t2)   // 没有成环就可以加进去啦
    {
        f[t2]=t1;
        flag=1;
    }
    return ;
}
int main()
{
    int i,j,t;
    while(~scanf("%d %d",&n,&m)&&(n||m))
    {
        int minn=inf;
        for(i=0; i<m; i++)
            scanf("%d %d %d",&list[i].a,&list[i].b,&list[i].l);
        sort(list,list+m,cmp);
        for(i=0; i<m; i++)
        {
            init();
            int num=0;
            for(j=i; j<m; j++)
            {
                merge(list[j].a,list[j].b);
                if(flag==1)
                {
                    num++;
                    if(num==n-1)  // 找到n-1条边就可以构成树
                    {
                        t=list[j].l-list[i].l;
                        if(t<minn)
                            minn=t;
                    }
                }
            }
        }
        if(minn==inf)
            printf("-1\n");
        else
            printf("%d\n",minn);
    }
    return 0;
}