POJ 3522 - Slim Span - [kruskal求MST]

题目连接：http://poj.org/problem?id=3522

Time Limit: 5000MS Memory Limit: 65536K

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 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 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

 

在全部生成树里，找到“最大边权值 减去 最小边权值”最小的那棵生成树。

那么，对于已经某个肯定的最小边的全部生成树，咱们找到最小生成树，它的“最大边权值 减去 最小边权值”就是这些生成树里最小的。

而后，咱们枚举最小边便可。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 #define N 102
 6 #define M 5000
 7 #define INF 2147483647
 8 int n,m;
 9 struct Edge{
10     int u,v,w;
11 }e[M];
12 bool cmp(Edge a,Edge b){return a.w<b.w;}
13 int par[N];
14 int find(int x){return( par[x]==x ? x : par[x]=find(par[x]) );}
15 int kruskal(int st)//得到最小边，做为开始边
16 {
17     int i,cnt=0;
18     for (i=1;i<=n;i++) par[i]=i;//初始化并查集 
19     for (i=st;i<m;i++)//遍历后面的每条边 
20     {
21         int x=find(e[i].u),y=find(e[i].v);
22         if (x != y){//若是这条边的链接的左右节点还未连通 
23             par[y]=x;//将这条边连通 
24             if (++cnt==n-1) break;//边计数增长1，若是边数到达了n-1条，那么一棵生成树已完成，跳出 
25         }
26     }
27     if (cnt<n-1) return -1; //若是从开始边日后遍历，遍历完了全部边，依然没法产生一颗生成树，那么返回-1 
28     return e[i].w-e[st].w; //不然就返回这棵生成树的“最大边权值 减去 最小边权值”的值 
29 }
30 int main()
31 {
32     while (scanf("%d%d",&n,&m) && n!=0)
33     {
34         for (int i=0;i<m;i++) scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
35         sort(e,e+m,cmp);//把边按权值按从小到大排序 
36         int tmp,ans=INF;
37         for (int i=0;i<m;i++)//枚举最小边 
38         {
39             tmp=kruskal(i);
40             if(tmp==-1) break;//若是从这条最小边开始已经没法产生生成树了，以后显然也不会有生成树了，那么咱们就直接跳出便可
41             if(tmp<ans) ans=tmp;//记录下最小的那个“最大边权值 减去 最小边权值”
42         }
43         if(ans==INF) printf("-1\n"); //若是答案没被更新过，那么显然连一棵生成树都没有，按题目要求打印-1 
44         else printf("%d\n",ans);//不然就打印出答案便可 
45     }
46     return 0;
47 }