Slim Span POJ - 3522

问题：

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 vbk connected 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：

让你构造生成树，使生成树最大权值与最小权值之差最小，问最小权值多大，如果不能构成一个生成树输出-1.

思路：

采用Kruskal算法，用sort从小到大排完序后，枚举最小边，形成生成树，保就下最小差。

代码：

#define N 110
#define inf 0x3f3f3f3f
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f[N],n,m;
struct node
{
    int u,v,w;
} k[N*N];
bool cmp(node x,node y)
{
    return x.w<y.w;
}
int getf(int a)
{
    if(f[a]==a)return a;
    return f[a]=getf(f[a]);
}
int merge(int a,int b)
{
    int t1=getf(a);
    int t2=getf(b);
    if(t1!=t2)
    {
        f[t2]=t1;
        return 1;
    }
    return 0;
}
int main()
{
    while(scanf("%d%d",&n,&m)&&(n||m))
    {
        for(int i=1; i<=m; i++)
            scanf("%d%d%d",&k[i].u,&k[i].v,&k[i].w);
        sort(k+1,k+m+1,cmp);
        int mi=0x3f3f3f3f;
        for(int h=1; h<=m; h++)//枚举每一条边为最小边
        {
            //Kruskal算法
            for(int i=0; i<=n; i++)
                f[i]=i;
            int countt=1;
            for(int i=h; i<=m; i++)//i=h是最小边之前的边都不在用
            {
                if(merge(k[i].u,k[i].v))
                {
                    countt++;
                }
                if(countt==n)
                {
                    mi=min(mi,k[i].w-k[h].w);//保留最小差
                    break;
                }
            }
        }
        if(mi<inf)
            printf("%d\n",mi);
        else
            printf("-1\n");
    }
    return 0;
}