poj 3522 Slim Span(枚举+最小生成树)

链接：

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

题目：

Slim Span

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4962		Accepted: 2587

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

Source

Japan 2007

题目大意：

一个图可能会构成多个生成树，那么求其中生成树中最小的一条边与最大的一条边之差最小的是多少。

分析与总结：

这题虽然不是求最小生成树，但是也很类似。 按照Kruskal的方法，把所有边排序，然后依次枚举最小边，开始构造生成树，取差值最小的便是。

代码：

#include
#include
using namespace std;

const int VN = 110;// 点的数量
const int EN = VN*VN; // 边的数量
const int INF = 10000000;
  
template
class Kruskal{
public:
    // n为点的数量
    void init(int _n){ 
        n=_n; size = 0;  make_set(n); MSTsum = 0;
    }
    void insert(int u, int v, Type w){
        if(u==v)return;
        E[size++].set(u,v,w);
    }
    Type kruskal(){
        sort(E, E+size);
        int minLen=INF;
        for(int i=0; irank[b]) f[b]=a;
        else{
            if(rank[a]==rank[b]) ++rank[b];
            f[a] = b;
        }
        return true;
    }
};

KruskalG;
int n, m;

int main(){
    int a,b,c;
    while(~scanf("%d%d",&n,&m)&&n+m){
        G.init(n);
        for(int i=0; i

——  生命的意义，在于赋予它意义。

          

     原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)