Slim Span
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 9841 Accepted: 5196
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
Source
Japan 2007
题意：求一个无向图的最大边和最小边差值最小的生成树。
思路：按照权值排序后暴力查找最小值和最大值。判断当前图是否联通。
更新最小值。
#include<iostream>
#include<algorithm>
#include<string.h>
#include<cstdio>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 300560;
struct node{
int u,v,cost;
}s[maxn];
int par[maxn];
int ran[maxn];
int n,m;
void init(int n){
for(int i=1;i<=n;i++){
par[i]=i;
ran[i]=1;
}
}
bool cmp(node a,node b){
return a.cost<b.cost;
}
int find(int x){
return x==par[x]?x:find(par[x]);
}
void unite(int x,int y){
x=find(x);
y=find(y);
if(x==y) return ;
if(ran[x]<ran[y]){
par[x]=y;
}
else{
par[y]=x;
if(ran[x]==ran[y]) ran[x]++;
}
}
bool same(int a,int b){
return find(a)==find(b);
}
bool jud(int n){
int res=0;
for(int i=1;i<=n;i++){
if(par[i]==i) res++;
}
return res==1;
}
int main(){
while(scanf("%d %d",&n,&m)&&(n+m)){
int ans=INF;
for(int i=1;i<=m;i++){
scanf("%d %d %d",&s[i].u,&s[i].v,&s[i].cost);
}
sort(s+1,s+m+1,cmp);
int flag=0;
for(int i=1;i<=m;i++){
init(n);
for(int j=i;j<=m;j++){
if(!same(s[j].u,s[j].v)){
unite(s[j].u,s[j].v);
if(jud(n)){
ans=min(ans,s[j].cost-s[i].cost);
flag=1;
}
}
}
}
if(!flag) puts("-1");
else printf("%d\n",ans);
}
}