Slim Span

Description
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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 .
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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).
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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.

nm
a1b1w1
⋮
ambmwm

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 vbk connected by the k -th 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
一个最小生成树的变形，把边从小到大排列，如果边加入不会形成环，那么就加进去，否则扔掉。
可以证明，如果边a小于边b，边b形成最小生成树，那么a加入就一定会形成环，这时候把b扔掉，形成的树的值就比之前的小，得证。
然后从最小边开始枚举使用KrusKal，使用完之后判断是否联通，如果联通就比较差值，最后就是答案了。

#include<cstdio>#include<cstring>#include<iostream>#include<cstdio>#include<cstring>#include<iostream>#include<queue>#include<vector>#include<algorithm>#include<string>#include<cmath>#include<set>using namespace std;typedef long long ll;int u[5005], v[5005], w[5005], r[5005], p[105], n, m;bool cmp(int a, int b){    return w[a] < w[b];}int find(int x){ return p[x] == x ? x : p[x] = find(p[x]); }bool check(){    int i, temp;    temp = find(1);    for (i = 2; i <= n; i++)    {        if (find(i) != temp)return false;    }    return true;}int KrusKal(int l){    int temp = 0, i, temp1, temp2;    temp1 = w[r[l]];    for (i = 1; i <= n; i++)p[i] = i;    for (i = l; i <= m; i++)    {        int e = r[i];        int x = find(u[e]);        int y = find(v[e]);        if (x != y)        {            temp2 = w[r[i]];            p[x] = y;        }    }    if (!check())    {        temp = 999999;    }    else        temp = temp2 - temp1;    return temp;}int main(){    int i, j, ans, t;    while (scanf("%d%d", &n, &m) != EOF)    {        if (!m&&!n)break;        for (i = 1; i <= m; i++)        {            scanf("%d%d%d", &u[i], &v[i], &w[i]);            r[i] = i;        }        sort(r + 1, r + 1 + m, cmp);        ans = 999999;        for (i = 1; i <= m; i++)        {            ans = min(ans, KrusKal(i));        }        if (ans == 999999)            cout << "-1" << endl;        else            cout << ans << endl;    }    return 0;}