POJ 3522-Slim Span（苗条树-kruskal生成树）

向右看齐 2022-06-16 03:22 137阅读 0赞

Slim Span

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;5000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;8025</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;4272</td> 
  </tr> 
 </tbody> 
</table>

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 \{ *v*1, *v*2, …, *vn*\} and *E* is a set of undirected edges \{ *e*1, *e*2, …, *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*.

![3522_1.png][]  
Figure 5: A graph *G* and the weights of the edges

For example, a graph *G* in Figure 5(a) has four vertices \{ *v*1, *v*2, *v*3, *v*4\} and five undirected edges \{ *e*1, *e*2, *e*3, *e*4, *e*5\}. The weights of the edges are *w*(*e*1) = 3, *w*(*e*2) = 5, *w*(*e*3) = 6, *w*(*e*4) = 6, *w*(*e*5) = 7 as shown in Figure 5(b).

![3522_2.png][]  
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.

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

Source

[Japan 2007][]

## 题目意思： ##

一棵生成树的最大边和最小边的差值最小时，称为苗条树，计算最小差值，不存在输出-1。

## 解题思路： ##

试着prim写了一半发现多棵生成树不好搞，换了kruskal，先对边排序，然后依次枚举，构造以每条边为最小边的生成树，计算更新差值。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <algorithm>
    #define MAXN 100010
    #define INF 0xfffffff
    using namespace std;
    int pre[MAXN],n,m,cnt;
    
    struct edge
    {
        int u,v,cost;
    } es[MAXN];
    bool cmp(edge e1,edge e2)
    {
        return e1.cost<e2.cost;
    }
    int set_find(int x)
    {
        if(pre[x]!=x)  return pre[x]=set_find(pre[x]);
        return x;
    }
    void unite(int p,int q)
    {
        p=set_find(p);
        q=set_find(q);
        if(p!=q) pre[p]=q;
    }
    bool same(int x,int y)
    {
        return set_find(x)==set_find(y);
    }
    void init_union_find(int n)
    {
        for(int i=0; i<=n; i++)
            pre[i]=i;
    }
    int kruskal(int src)
    {
        init_union_find(n);//并查集初始化
        int cnt=0;
        for(int i=src; i<m; ++i)
        {
            edge e=es[i];
            if(!same(e.u,e.v))//不在一棵树中
            {
                unite(e.u,e.v);//合并
                ++cnt;//更新边的数量
            }
            if(cnt==n-1) return (es[i].cost-es[src].cost);//构成一棵生成树
        }
        return INF;
    }
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("G:/cbx/read.txt","r",stdin);
        //freopen("G:/cbx/out.txt","w",stdout);
    #endif
        ios::sync_with_stdio(false);
        cin.tie(0);
        while(cin>>n>>m)
        {
            if(n==0&&m==0) break;
            int ans=INF;
            for(int i=0; i<m; ++i)
                cin>>es[i].u>>es[i].v>>es[i].cost;
            sort(es,es+m,cmp);//按照edge.cost的顺序升序排列
            for(int i=0; i<=m-n+1; ++i)//枚举每条边作为最小边的
            {
                int res=kruskal(i);//当前最大边和最小边的差值
                ans=min(ans,res);//更新最小答案
            }
            if(ans==INF) ans=-1;
            cout<<ans<<endl;
        }
        return 0;
    }

[3522_1.png]: /images/20220616/ab90b912726644a0a358448400162ca9.png
[3522_2.png]: /images/20220616/f11ef60c3ebf4d83939bedff6f5dd55b.png
[Japan 2007]: http://poj.org/searchproblem?field=source&key=Japan+2007