UVA 1395 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, ... , v_n} and E is a set of undirected edges {e₁, e₂, ... , e_m}. 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 T_a 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 T_a is 4. The slimnesses of spanning trees T_b , T_c and T_d 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 T_d 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. a_k and b_k (k = 1, ... , m) are positive integers less than or equal to n, which represent the two vertices v_ak and v_bk connected by the kth edge e_k. w_k is a positive integer less than or equal to 10000, which indicates the weight of e_k . 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

Output for the Sample Input

1
20
0
-1
-1
1
0
1686
50

总结：

一开始认为是另一种标准的生成树，准备重新搞一套选择边的方式，（虽然最后用暴力枚举的方法，但是更优的策略未必不存在），想了半天也没想出怎么搞，但是期间脑海里确实是有闪过，先进性排序，如果直接从小到大的生成树，活着从大到小的生成树，或者从中间开始的生成树可不可以呢，因为这样都是从相邻的中选择的，但是也只是闪过了一下，没有继续思考下去，看过书上的提示以后才发现其中的这种思想的本质就是在枚举答案啊，因为我们每次选择的区间［l,r］那么我们的答案就是w[r]-w[r]啊，好痛心啊，以前只会枚举的时候看见什么都想暴力枚举，现在学的东西越来越多了反倒想不到枚举了！

//
//  main.cpp
//  Slim Span
//
//  Created by 张嘉韬 on 16/8/20.
//  Copyright © 2016年 张嘉韬. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=100+10;
const int inf=1<<30;
int n,m,f[maxn],minimum,maximum;
struct road
{
    int u;
    int v;
    int w;
}roads[maxn*maxn/2];
int cmp(road a,road b)
{
    return a.w<b.w;
}
int getf(int i)
{
    if(f[i]==i) return i;
    else return f[i]=getf(f[i]);
}
void merge(int a,int b,int w)
{
    maximum=max(maximum,w);
    minimum=min(minimum,w);
    int fa=getf(a);
    int fb=getf(b);
    if(fa==fb) return;
    f[fa]=fb;
}
void init()
{
    for(int i=1;i<=n;i++) f[i]=i;
    minimum=inf;
    maximum=-inf;
}
int check()
{
    int cnt=0;
    int father=getf(1);
    for(int i=2;i<=n;i++)
    {
        if(getf(i)!=father) cnt++;
    }
    if(cnt==0) return 1;
    else return 0;
}
int main(int argc, const char * argv[]) {
    //freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin);
    while(scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0) break;
        for(int i=1;i<=m;i++) scanf("%d%d%d",&roads[i].u,&roads[i].v,&roads[i].w);
        sort(roads+1,roads+m+1,cmp);
        int silm=inf;
        for(int l=1;l<=m;l++)
        {
            init();
            for(int r=l;r<=m;r++)
            {
                if(getf(roads[r].u)!=getf(roads[r].v))
                {
                    merge(roads[r].u,roads[r].v,roads[r].w);
                    if(check()==1)
                    {
                        if(maximum-minimum<silm) silm=maximum-minimum;
                        break;
                    }
                }
            }
        }
        if(silm==inf) printf("-1\n");
        else printf("%d\n",silm);
    }
    return 0;
}