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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).
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 kth 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：求生成树中最大边与最小边的差值的绝对值最小的一个。要知道差值就需要知道生成树的最大边和最小边，我们知道当我们用克鲁斯卡尔求最小生成树的时候，每次都是找最小边，构成的最小生成树的所有边都是最小的。如果我们遍历所有边，再在该边的基础上用克鲁斯求最小生成树，那么求得的最大边与该边的差值肯定最小。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int INF = 0x3fffffff;
struct Node
{
int from;
int to;
int cost;
bool operator< (Node t) const
{
return cost < t.cost;
}
};
Node e[10004];
int pre[102];
int ans;
int find(int x)
{
return x == pre[x] ? x : pre[x] = find(pre[x]);
}
void kruskal(int n,int k)
{
sort(e,e + k);
int cnt = 0;
ans = INF;
for(int i = 0;i < k;i++)
{
cnt = 0;
for(int j = 1;j <= n;j++)
{
pre[j] = j;
}
for(int j = i;j < k;j++)
{
int x = find(e[j].from);
int y = find(e[j].to);
if(x == y)
{
continue;
}
pre[x] = y;
if(++cnt == n - 1)
{
ans = min(ans,e[j].cost - e[i].cost);
break;
}
}
}
if(ans == INF)
{
ans = -1;
}
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m) != EOF && (n + m) != 0)
{
int u,v,c;
for(int i = 0;i < m;i++)
{
scanf("%d%d%d",&u,&v,&c);
e[i].from = u;
e[i].to = v;
e[i].cost = c;
}
kruskal(n,m);
printf("%d\n",ans);
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。
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原文地址：http://blog.csdn.net/wang2534499/article/details/47843161
