POJ 3522 Slim Span（生成树+克鲁斯卡尔）

女爷i 2022-06-10 02:52 149阅读 0赞

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*.

![4c86d7c20f5838ac2bc7cda82ca110a6_v_1501863254][]  
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).

![7668f083800f87e7390a284fccd2d61d_v_1501863254][]  
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

题解：

题意：

给一堆边，让你求一个生成树的最大边-最小边的最小值是多少，如果不能建树输出-1

思路：

遍历所有可能的生成树来求，就用克鲁斯卡尔算法，先将边按照值从小到大排序，然后从0开始建树，从1开始建树。。直到无法建树，求加入并查集的最后一条边和第一条边的差  
代码：

#include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<stdio.h>
 #include<math.h>
 #include<string>
 #include<stdio.h>
 #include<queue>
 #include<stack>
 #include<map>
 #include<deque>
 #define M (t[k].l+t[k].r)/2
 #define lson k*2
 #define rson k*2+1
 #define ll long long
 using namespace std;
 const int INF=100861111;
 int pre[105],n,m,tag,minn;
 struct edge
 {
 int f,t;
 int v;
 }a[10005];
 int cmp(edge x,edge y)//边从小到大排序
 {
 return x.v<y.v;
 }
 int find(int x)//并查集
 {
 if(x!=pre[x])
 pre[x]=find(pre[x]);
 return pre[x];
 }
 void klske(int start)//克鲁斯卡尔算法，从边start开始
 {
 int i,j,ans=0,d1,d2;
 for(i=1;i<=n;i++)//初始化并查集
 pre[i]=i;
 for(i=start;i<m;i++)//克鲁斯卡尔
 {
 d1=find(a[i].f);
 d2=find(a[i].t);
 if(d1!=d2)
 {
 pre[d2]=d1;
 ans++;
 if(ans>=n-1)
 {
 tag=1;
 if(minn>a[i].v-a[start].v)//最后一条边减去第一条边
 minn=a[i].v-a[start].v;
 return;
 }
 }
 }
 }
 int main()
 {
 int i,j;
 while(scanf("%d%d",&n,&m)!=EOF)
 {
 if(n==0&&m==0)
 break;
 for(i=0;i<m;i++)
 {
 scanf("%d%d%d",&a[i].f,&a[i].t,&a[i].v);
 }
 sort(a,a+m,cmp);
 tag=0;
 minn=INF;
 for(i=0;i<=m-n+1;i++)//最多从m-n+1开始建树，因为至少要n-1条边
 {
 klske(i);
 if(!tag)//如果不能建树
 break;
 }
 if(!tag)
 printf("-1\n");
 else
 printf("%d\n",minn);
 }
 return 0;
 }

[4c86d7c20f5838ac2bc7cda82ca110a6_v_1501863254]: https://odzkskevi.qnssl.com/4c86d7c20f5838ac2bc7cda82ca110a6?v=1501863254
[7668f083800f87e7390a284fccd2d61d_v_1501863254]: https://odzkskevi.qnssl.com/7668f083800f87e7390a284fccd2d61d?v=1501863254