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, … , vn} and E is a
set of undirected edges {e1, e2, … , em}. 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).
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 Ta 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 Ta is 4. The slimnesses of spanning trees Tb, Tc
and Td 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. ak and bk (k = 1, … , m) are positive integers less than or equal to n, which
represent the two vertices vak
and vbk
connected by the k-th edge ek. wk is a positive integer less than
or equal to 10000, which indicates the weight of ek. 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
题解：
枚举L，当所有边连通的时候，肯定存在一个最小的生成树使W[R]-W[L]最小。
从小到大枚举L，对于每个L，进行Kruskal算法
CODE
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int u[10010],v[10010],w[10010],r[10010];
int p[10010];
int m,n;
const int INF = 0x3f3f3f3f;
int cmp(const int i, const int j)
{
return w[i] < w[j];
}
int find(int x)
{
return p[x] == x ? x : p[x] = find(p[x]);
}
int Kruskal()
{
int ans = INF;
for(int i=0; i<m; i++) r[i] = i;
sort(r,r+m,cmp);
for(int j=0; j<m; j++)
{
int cnt = 1;
for(int i=1; i<=n; i++) p[i] = i;//从点1开始
int flag = 1;
for(int i=j; i<m; i++)
{
int e = r[i];
int x = find(u[e]);
int y = find(v[e]);
if(x != y) {
cnt++;
p[x] = y;
}
if(cnt == n) {
ans = min(ans,w[r[i]]-w[r[j]]);
flag = 0;
break;
}
}
if(flag) break;
}
if(ans == INF) ans = -1;
return ans;
}
int main()
{
while(cin >> n >> m && m || n)
{
for(int i=0; i<m; i++)
cin >> u[i] >> v[i] >> w[i];
cout << Kruskal() << endl;
}
return 0;
}