poj 3522 Slim Span (最小生成树kruskal)

http://poj.org/problem?id=3522

Slim Span

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5666		Accepted: 2965

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.

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 andb_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 ofe_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

Sample Output

1
20
0
-1
-1
1
0
1686
50

Source

Japan 2007

 

【题解】：

　这道题简单的来说就是求一棵生成树使最大的边和最小的边差值最小。

　　换个角度想就是用n-1条（n个点）数值相差不多的边，组成一棵生成树。 在生成树的prim和kruskal两个算法中很容易就会觉得kruskal的贪心思想会更加适合这道题。 kruskal算法一开始会对边进行排序，然后枚举最小的边。

【code】：

 /**
 Judge Status:Accepted    Memory:756K
 Time:157MS     Language:G++
 Code Lenght:1613B  Author:cj
 */

 #include<iostream>
 #include<stdio.h>
 #include<string.h>
 #include<algorithm>

 #define N 110
 #define M 6000
 #define INF 1000000000

 using namespace std;

 struct Nod
 {
     int a,b,c;
 }node[M];

 int n,m,parent[N];

 bool cmp(Nod a,Nod b)
 {
     return a.c<b.c;
 }

 int findp(int a)
 {
     while(a!=parent[a])
     {
         a=parent[a];
     }
     return a;
 }

 int merge(Nod nd)
 {
     int x = findp(nd.a);
     int y = findp(nd.b);
     if(x!=y)
     {
         parent[x]=y;
         return nd.c;
     }
     return -;
 }

 int kruskal(int id)
 {
     int i,cnt = ;
     if(m-id+<n-)  return INF;  //少于n-1边的话 注定够不成生成树
     for(i=;i<=N;i++)   parent[i]=i;
     int flag = ,mins
     for(i=id;i<m;i++)
     {
         int temp = merge(node[i]);
         if(temp!=-)
         {
             if(!flag)   mins = temp;  //记录最小边
             flag = ;
             cnt++;
         }
         if(cnt>=n-)    return temp-mins;  //只要找到n-1条边即可，返回最大边与最小边的差
     }
     if(cnt<n-) return INF;  //构不成生成树
 }

 int main()
 {

     while(~scanf("%d%d",&n,&m))
     {
         if(n==&&m==)  break;
         if(m==){puts("-1");continue;}
         int i;
         for(i=;i<m;i++)
         {
             scanf("%d%d%d",&node[i].a,&node[i].b,&node[i].c);
         }
         sort(node,node+m,cmp);
         int ans = INF;
         int temp = kruskal();
         if(temp==INF)
         {
             puts("-1");
             continue;
         }
         if(ans>temp)    ans = temp;
         for(i=;i<m;i++)
         {
             temp = kruskal(i);  //枚举最小边
             if(ans>temp)    ans = temp;
         }
         printf("%d\n",ans);
     }
     return ;
 }