POJ 3522 - Slim Span - [kruskal求MST]

題目連接：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

 

在全部生成樹裏，找到「最大邊權值 減去 最小邊權值」最小的那棵生成樹。

那麼，對於已經某個肯定的最小邊的全部生成樹，咱們找到最小生成樹，它的「最大邊權值 減去 最小邊權值」就是這些生成樹裏最小的。

而後，咱們枚舉最小邊便可。

 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)//得到最小邊，做爲開始邊
16 {
17     int i,cnt=0;
18     for (i=1;i<=n;i++) par[i]=i;//初始化並查集 
19     for (i=st;i<m;i++)//遍歷後面的每條邊 
20     {
21         int x=find(e[i].u),y=find(e[i].v);
22         if (x != y){//若是這條邊的鏈接的左右節點還未連通 
23             par[y]=x;//將這條邊連通 
24             if (++cnt==n-1) break;//邊計數增長1，若是邊數到達了n-1條，那麼一棵生成樹已完成，跳出 
25         }
26     }
27     if (cnt<n-1) return -1; //若是從開始邊日後遍歷，遍歷完了全部邊，依然沒法產生一顆生成樹，那麼返回-1 
28     return e[i].w-e[st].w; //不然就返回這棵生成樹的「最大邊權值 減去 最小邊權值」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);//把邊按權值按從小到大排序 
36         int tmp,ans=INF;
37         for (int i=0;i<m;i++)//枚舉最小邊 
38         {
39             tmp=kruskal(i);
40             if(tmp==-1) break;//若是從這條最小邊開始已經沒法產生生成樹了，以後顯然也不會有生成樹了，那麼咱們就直接跳出便可
41             if(tmp<ans) ans=tmp;//記錄下最小的那個「最大邊權值 減去 最小邊權值」
42         }
43         if(ans==INF) printf("-1\n"); //若是答案沒被更新過，那麼顯然連一棵生成樹都沒有，按題目要求打印-1 
44         else printf("%d\n",ans);//不然就打印出答案便可 
45     }
46     return 0;
47 }