POJ

Slim Span

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.

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 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

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题目大意：
对于一个无向图来说，它可能有多个生成树， 定义一个生成树的slimness为这个树的边的最大权值减去最小权值。试图求出一个给定图中所有生成树中最小的slimness值。

解题思路：
1.对于kruskal算法来说，其各个边是按照大小顺序依次加入的，所以slimness值就是最后加入的边长减去第一个加入的变边长。
2.对于同一图来说，最小生成树可能有多个，但是其各个边的权值一定是相等的。也就是不存在这样的a+b = c+d，其中a,b为最小生成树x的边 的权值而c,d为最小生成树y的边的权值，而x,y均为同一图的最小生成树。
3.继而推广，对于一个图的最小生成树的最小边权值，其唯一对应一个最大边的权值。
4.回到这道题，我们先将所有的边全部记录，然后枚举最小边的边长，对于每一个最小边长做一次Kruskal计算出silmness值。计算完当前边以后，下一次计算将跳到下一个有更大的边值的边。
5.枚举结束的条件可以设置多个：1.silmness值已经到0   2.从起始位置的边到最大的边，剩余数量已经不足n-1（构不成生成树了）  3.上次Kruskal计算中，已经求不出一个最小生成树（后续的图相当于前面求的图的子图，所以也将求不出最小生成树）。

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源代码：

#include<iostream>#include<stdio.h>#include<vector>#include<algorithm>#include<string.h>using namespace std;int n,e;int fa[105];struct edg{    int start,end;    int cost;    bool operator < (const edg& a) const{        return cost<a.cost;    }}; edg roads[10000];int find(int x){    int r=x;    while(r!=fa[r])        r=fa[r];    int i=x,j;    while(fa[i]!=r){        j=fa[i];        fa[i]=r;        i=j;    }    return r;}  int kruskal(int start_pos){    for(int i=0;i<=n;i++)      fa[i] = i;    int now_choose=0;    for(int i=start_pos;i<e;i++){        int fs = find(roads[i].start);        int fe = find(roads[i].end);                if(fs!=fe){            fa[fs] = fe;            now_choose++;        }               if(now_choose==n-1){            return roads[i].cost - roads[start_pos].cost;        }    }    return -1;}int main(){    while(scanf("%d%d",&n,&e)){        if(n==0 && e==0)          break;        for(int i=0;i<e;i++){            scanf("%d%d%d",&roads[i].start,&roads[i].end,&roads[i].cost);        }        sort(roads,roads+e);        int start_pos = 0,last_len = -1,ans=1e9;        while(ans  &&  e-start_pos>=n-1){            int now_res = kruskal(start_pos);            if(now_res>=0)              ans = min(ans,now_res);            else              break;             last_len = roads[start_pos].cost;            while(roads[start_pos].cost == last_len)              start_pos++;        }        ans = ans==1e9?-1:ans;        printf("%d\n",ans);    }    return 0;}