POJ3522 Slim Span(kruskal)

Slim Span

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 8213 Accepted: 4375

Description

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 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 ⋮ a_mb_mw_m

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 51 2 31 3 51 4 62 4 63 4 74 61 2 101 3 1001 4 902 3 202 4 803 4 402 11 2 13 03 11 2 13 31 2 22 3 51 3 65 101 2 1101 3 1201 4 1301 5 1202 3 1102 4 1202 5 1303 4 1203 5 1104 5 1205 101 2 93841 3 8871 4 27781 5 69162 3 77942 4 83362 5 53873 4 4933 5 66504 5 14225 81 2 12 3 1003 4 1004 5 1001 5 502 5 503 5 504 1 1500 0

Sample Output

1200-1-110168650

要求一个满足条件（最细）的生成树，如果没有生成树输出-1

对次我们就要枚举生成树，但暴力枚举肯定不行，既然与边的最大值最小值有关，就想到kruskal

对于一个图来说，最小生成树不是唯一的，但最小生成树的

最小生成树的性质， 一个图的最小生成树不一定是唯一的，但是组成这些最小生成树的各个边的权值一定都是一一对应相同的

所以对于每一个最小值，形成的最小生成树都有唯一的最大值，因此每次删去最小边，求最小生成树，就能考虑所有的情况

#include <iostream>#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;const int N=105;int f[N],n,m,flag,ans;struct node{    int u,v,w;}e[N*N];int getf(int x){    if(f[x]==x)return x;    f[x]=getf(f[x]);    return f[x];}bool mix(int x,int y){    int tx=getf(x);    int ty=getf(y);    if(tx!=ty)    {        f[tx]=ty;        return true;    }    return false;}bool cmp(node a,node b){    return a.w<b.w;}void init(){    for(int i=1;i<=n;i++)        f[i]=i;}void kruskal(int x){    init();    int cnt=0,minn=0x3f3f3f3f,maxx=0;    for(int i=x;i<=m;i++)    {        if(mix(e[i].u,e[i].v))        {            minn=min(e[i].w,minn);            maxx=max(e[i].w,maxx);            cnt++;        }        if(cnt==n-1){flag=1;ans=min(ans,maxx-minn);break;}    }}int main(){    while(~scanf("%d%d",&n,&m)&&n+m)    {        flag=0,ans=0x3f3f3f3f;        for(int i=1;i<=m;i++)            scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);            sort(e+1,e+1+m,cmp);           for(int i=1;i<=m;i++)           {               kruskal(i);           }           if(!flag)printf("-1\n");           else        printf("%d\n",ans);    }    return 0;}