POJ 3522 Slim Span "enum + Kruskal minimum spanning tree"

Source: Internet
Author: User


Slim Span

Time Limit: 5000MS

Memory Limit: 65536K

Total Submissions: 7365

Accepted: 3909

Description

Given an undirected weighted graph G, you should find one of spanning trees specified as follows.

The graph 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 ee have 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 Weig HT 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) have 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 is several spanning trees for G. Four of them is depicted in Figure 6 (a) ~ (d). The Spanning Tree Ta in Figure 6 (a) had three edges whose weights are 3, 6 and 7. The largest weight is 7 and the smallest weight are 3 so, 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) is 3, 2 and 1 , respectively. You can easily see the slimness of any other spanning tree are greater than or equal to 1, thus the spanning tree Td in Fig Ure 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 and 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 is 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.& nbsp AK  and BK   ( k  = 1, ...,   m ) is positive integers less than or equal to  n , which represent the vertices  vak  and  vbk  connected by the  k th edge  ek .   wk  is a positive integer less than or equal to 10000, WHI CH indicates the weight of ek . You can assume this graph  G  = ( V ,   E ) is easy, that's, there is no self-l Oops (that connect to the same vertex) nor parallel edges (that is, or more edges whose both ends is the same, vertic ES).

Output

For each dataset, if the graph has spanning trees, the smallest slimness among them should is 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

The main topic: give you n points, M edge, to find a spanning tree, so that its maximum weight of the edge-the minimum weight of the weight of the minimum value.


Analysis of the topic: Survival into a tree, so that the maximum weight of the edge-the minimum weight of the weight of the edge as small as possible, we can not be difficult to think of using Kruskal greedy to find the minimum spanning tree algorithm, because the greedy to make the maximum weight of the edge-minimum weight margin as small as possible, but because it is not the smallest spanning So we use the enumeration method to solve, the point is not much, back to the 5000ms. Of course, you can use the enumeration method to solve the problem.

AC Code:


#include <stdio.h> #include <iostream> #include <algorithm> #include <string.h>using namespace std;struct path{int x,y,w;} a[100000];int f[10000];int n,m;int output;int cmp (path A,path b) {return A.W&LT;B.W;}    int find (int a) {int r=a;    while (F[R]!=R) r=f[r];    int i=a;    Int J;        while (i!=r) {j=f[i];        F[i]=r;    I=j; } return R;}    void merge (int a,int b) {int A, B;    A=find (a);    B=find (b); if (a!=b) f[b]=a;}    void Init () {for (int i=1; i<=n; i++) {f[i]=i;    }}void solve (int x) {int cont=0;    int minn=0x1f1f1f1f;    int maxn=-0x1f1f1f1f;    Init ();            for (int i=x; i<m; i++) {if (Find (a[i].x)!=find (a[i].y)) {minn=min (A[i].w,minn);            Maxn=max (A[I].W,MAXN);            Merge (A[I].X,A[I].Y);        cont++;        }} if (cont==n-1) {//printf ("%d\n", Maxn-minn);    Output=min (Output,maxn-minn); }}int Main () {while (~scanf ("%d%d", &AMP;N,&AMp;m)) {if (n+m==0) break;        output=0x1f1f1f1f;        for (int i=0; i<m; i++) {scanf ("%d%d%d", &AMP;A[I].X,&AMP;A[I].Y,&AMP;A[I].W);        } sort (a,a+m,cmp);            for (int i=0, i<m; i++) {if (m-i<n-1) break;        Solve (i);        }//printf ("%d\n", output);        if (output!=0x1f1f1f1f) printf ("%d\n", output);    else printf (" -1\n"); }}









POJ 3522 Slim Span "enum + Kruskal minimum spanning tree"

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