Our lovely KK has a difficult Social problem.
A big earthquake happened in his area.
N(2≤N≤2000)N(2≤N≤2000) cities have been implicated. All the roads between them are destroyed.
Now KK was commissioned to rebuild these roads.
However, after investigation，KK found that some roads are too damaged to rebuild.
Therefore, there are only M(0≤M≤15000)M(0≤M≤15000) roads can be rebuilt.
KK needs to make sure that there is a way between any two cities, and KK wants to rebuild roads as few as possible.
With rebuilding minimal number of roads, he also wants to minimize the difference between the price of the most expensive road been built and the cheapest one.
Input
The first line of the input file contains an integer T(1≤T≤10)T(1≤T≤10), which indicates the number of test cases.
For each test case,The first line includes two integers N(2≤N≤2000)N(2≤N≤2000)，M(0≤M≤15000)M(0≤M≤15000).
The next MM lines include three integers a,b,c(a≠b,1≤c≤2∗109)a,b,c(a≠b,1≤c≤2∗109),indicating there is a undirected edge between aa and bb,the cost is cc.
Output
For each test case, output the smallest difference between the price of the most expensive road and the cheapest one.If there is no legal solution, then output -1.
Sample Input
2 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 2 0
Sample Output
1686 -1
题解：这道题的时间给的是6s，排完序后所以可以枚举每一种修路的情况，但是如果把所有的情况都存在数组里会溢出，就只把最贵的和最便宜的放数组然后做差，然后找到最小的。
此题坑点有主要有二：
1，存所有情况数组溢出
2，定义那个较大的数一定要尽量大（在比较求差值最小的时候）
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int pre[100005];
struct node
{
int x,y;
long long int val;
}road[200005];
int find(int x)
{
if(x==pre[x])
return x;
else
{
return pre[x]=find(pre[x]);
}
}
bool merge(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
{
pre[fx]=fy;
return true;
}
else
{
return false;
}
}
bool cmp(node x,node y)
{
return x.val<y.val;
}
int main()
{
int n;
cin>>n;
while(n--)
{
int m,s;
cin>>m>>s;
for(int j=1;j<=m;j++)
{
pre[j]=j;
}
for(int t=1;t<=s;t++)
{
scanf("%d%d%d",&road[t].x,&road[t].y,&road[t].val);
}
sort(road+1,road+s+1,cmp);
long long int cnt=0;
int a[3000005];
int crt=0;
int l=0;
for(int t=1;t<=s;t++)
{
for(int j=t;j<=s;j++)
{
if(merge(road[j].x,road[j].y))
{
if(cnt%(m-1)==m-2||(cnt%(m-1)==0))
{
a[crt]=road[j].val;
crt++;
}
cnt++;
}
if(cnt%(m-1)==0)
{
for(int k=1;k<=m;k++)
{
pre[k]=k;
}
break;
}
}
}
long long int Min=999999999;
int k=0;
if(crt%2==1){
for(int t=0;t<crt-1;t=t+2)
{
if(Min>a[t+1]-a[t])
{
Min=a[t+1]-a[t];
}
}
}
else
{
for(int t=0;t<crt;t=t+2)
{
if(Min>a[t+1]-a[t])
{
Min=a[t+1]-a[t];
}
}
}
if(cnt>=m-1)
{
printf("%lld
",Min);
}
else
{
printf("-1
");
}
}
return 0;
}