poj3522最小生成树



Slim Span

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 8740 Accepted: 4659

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} andE is a set of undirected edges {e₁,e₂, …, e_m}. Each edgee ∈ E has its weight w(e).

A spanning tree T is a tree (a connected subgraph without cycles) which connects all the n vertices withn − 1 edges. The slimness of a spanning tree T is defined as the difference between the largest weight and the smallest weight among then − 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 arew(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 treeT_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 treeT_a is 4. The slimnesses of spanning treesT_b, T_c andT_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 andb_k (k = 1, …, m) are positive integers less than or equal to n, which represent the two verticesv_ak andv_bk connected by thekth edge e_k. w_k is a positive integer less than or equal to 10000, which indicates the weight ofe_k. You can assume that the graphG = (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

1200-1-110168650

题意是求所有的生成树中最长的边和最小边的最小差距。把所有的树都遍历一遍然后在找最小的差距。

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#define inf 0x3f3f3fusing namespace std;struct node{    int x,y,z;} no[10005];int f[105];int fin(int x){    if(x!=f[x])        return f[x]=fin(f[x]);    else        return x;}int w(int x,int y){    int p=fin(x);    int q=fin(y);    if(p!=q)    {        f[p]=q;        return 1;    }    else        return 0;}void in(){    for(int i=0; i<105; i++)        f[i]=i;}bool cmp(node m,node n){    return m.z<n.z;}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        if(m==0&&n==0)            break;        memset(no,0,sizeof(no));        for(int i=0; i<m; i++)        {            scanf("%d%d%d",&no[i].x,&no[i].y,&no[i].z);        }        sort(no,no+m,cmp);        in();        int l=0;        int ans=inf;        for(int i=0; i<m; i++)        {            int k=i;            int flag=0;            if(k>m-n+1)                break;            int cnt=0;            int ma=-inf,mi=inf;            for(int j=k; j<m; j++)            {                if(w(no[j].x,no[j].y)==1)                {                    cnt++;                    ma=max(ma,no[j].z);                    mi=min(mi,no[j].z);                }                if(cnt==n-1)                {                    flag=1;                    l=1;                    break;                }            }            in();            if(flag==1)                ans=min(ans,(ma-mi));        }        if(l==0)            printf("-1\n");        else            printf("%d\n",ans);    }}