Description
Given an undirected weighted graph G<tex2html_verbatim_mark> , you should find one of spanning trees specified as follows.
The graph G<tex2html_verbatim_mark> is an ordered pair (V, E)<tex2html_verbatim_mark> , where V<tex2html_verbatim_mark> is a set of vertices {v1, v2,..., vn}<tex2html_verbatim_mark> and E<tex2html_verbatim_mark> is a set of undirected edges {e1, e2,..., em}<tex2html_verbatim_mark> . Each edge e
A spanning tree T<tex2html_verbatim_mark> is a tree (a connected subgraph without cycles) which connects all the n<tex2html_verbatim_mark> vertices with n - 1<tex2html_verbatim_mark> edges. The slimness of a spanning tree T<tex2html_verbatim_mark> is defined as the difference between the largest weight and the smallest weight among the n - 1<tex2html_verbatim_mark> edges of T<tex2html_verbatim_mark> .
<tex2html_verbatim_mark>For example, a graph G<tex2html_verbatim_mark> in Figure 5(a) has four vertices {v1, v2, v3, v4}<tex2html_verbatim_mark> and five undirected edges {e1, e2, e3, e4, e5}<tex2html_verbatim_mark> . The weights of the edges are w(e1) = 3<tex2html_verbatim_mark> , w(e2) = 5<tex2html_verbatim_mark> , w(e3) = 6<tex2html_verbatim_mark> , w(e4) = 6<tex2html_verbatim_mark> , w(e5) = 7<tex2html_verbatim_mark> as shown in Figure 5(b).
<tex2html_verbatim_mark>There are several spanning trees for G<tex2html_verbatim_mark> . Four of them are depicted in Figure 6(a)∼(d). The spanning tree Ta<tex2html_verbatim_mark> 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<tex2html_verbatim_mark> is 4. The slimnesses of spanning trees Tb<tex2html_verbatim_mark> , Tc<tex2html_verbatim_mark> and Td<tex2html_verbatim_mark> 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<tex2html_verbatim_mark> 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<tex2html_verbatim_mark>m<tex2html_verbatim_mark>
a1<tex2html_verbatim_mark>b1<tex2html_verbatim_mark>w1<tex2html_verbatim_mark>
am<tex2html_verbatim_mark>bm<tex2html_verbatim_mark>wm<tex2html_verbatim_mark>
Every input item in a dataset is a non-negative integer. Items in a line are separated by a space.
n<tex2html_verbatim_mark> is the number of the vertices and m<tex2html_verbatim_mark> the number of the edges. You can assume 2
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
抽象：在數組中找出n-1個數，讓這n-1個數的最大值和最小值的差最小，先排序然後從第一個位置開始列舉即可，到 m - n+1結束
最小生成樹用克魯斯卡爾做的時候建議pre裏從0開始，我是從1開始的，wa了兩次，這是因爲我枚舉邊的時候多枚舉了一次，導致出現了0,會出現隨機數導致的誤差出現
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int pre[110];
struct node{
int x,y,values;
}edges[10010];
int cmp(node a ,node b){
return a.values < b.values;
}
void init(){
for(int i = 1;i<=110;i++){
pre[i] = i;
}
}
int finds(int x){
while(x != pre[x]){
x = pre[x];
}
return x;
}
int main(){
int n,m;
while(scanf("%d%d",&n,&m),n||m){
int a,b,c;
int minn = 0x3f3f3f;
for(int i = 0 ;i<m ;i++){
scanf("%d%d%d",&a,&b,&c);
edges[i].x = a;
edges[i].y = b;
edges[i].values = c;
}
sort(edges,edges+m,cmp);
for(int i = 0;i<= m-n +1;i++){
//建樹並且得出最小的值，如果建樹的過程中存在到達最後一條邊了，則結束循環
int j = i;
int nums = 0;
init();
for(;j<m;j++){
int u = finds(edges[j].x);
int v = finds(edges[j].y);
if(u != v){
pre[u] = v;
nums ++;
}
if(nums == n-1){
minn = min(minn,edges[j].values - edges[i].values);
break;
}
}
if(j == m )break;
}
printf("%d\n",minn == 0x3f3f3f? -1:minn);
}
}