uva 1395 瓶颈树

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 .

 

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).

 

=6in

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 T_d 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

a_m b_m w_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 2n100 and 0mn(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 k -th 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

参考了大神题解

因为kruskal求出的mst中最大边权最小，所以枚举最小边权，然后求差值即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>

using namespace std;

#define LL long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define cint const int

#define INF 100000000
#define MAXN 111
#define MAXM 10000

struct edge{
    int u, v, w;
    bool operator < (const edge &rhs)const{
        return w < rhs.w;
    }
}e[MAXM];
int p[MAXN], n, m;

int finds(int x){
    if(p[x]==-1) return x;
    return (p[x]=finds(p[x]));
}

bool mst(int k, int &dif){
    int i, j;
    fill_n(p+1, n, -1);     dif=0;
    for(i=k, j=0; i<m; i++){
        int fx = finds(e[i].u), fy = finds(e[i].v), w = e[i].w;
        if(fx!=fy){
            p[fx] = fy;
            dif = max(dif, w);
            if(++j == n-1){
                dif = dif - e[k].w;
                return true;
            }
        }
    }
    return false;
}

int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    while(scanf(" %d %d", &n, &m)==2 && n){
        int i, u, v, w;
        for(i=0; i<m; i++){
            scanf(" %d %d %d", &u, &v, &w);
            e[i]=(edge){u, v, w};
        }
        sort(e, e+m);
        int ans = INF, dif;
        for(i=0; i<m; i++){
            if(mst(i, dif)) ans = min(ans, dif);
            else break;
        }
        if(ans == INF) ans=-1;
        printf("%d\n", ans);
    }
    return 0;
}