Problem Description
There is a complete graph containing n vertices, the weight of the i-th vertex is wi.
The length of edge between vertex i and j (i≠j) is ⌊√|wi−wj|⌋.
Calculate the length of the shortest path from 1 to n.
Input
The first line of the input contains an integer T (1≤T≤10) denoting the number of test cases.
Each test case starts with an integer n (1≤n≤105) denoting the number of vertices in the graph.
The second line contains n integers, the i-th integer denotes wi (1≤wi≤105).
Output
For each test case, print an integer denoting the length of the shortest path from 1 to n.
Sample Input
1
3
1 3 5
Sample Output
2
题意：给出n个点的完全图，边的权值为两点绝对值之差的开根号，问从1走到n的最短路花费。
思路：容易证明 ⌊√a⌋+⌊√b⌋≥⌊√a+b⌋，进而可以证明边权满足三角不等式，故直接从 1走到 n就是最优的。
const int maxn=100010;
int n,m,k,a[maxn];
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
int k=abs(a[n-1]-a[0]);
k=sqrt(k);
printf("%d\n",k);
}
return 0;
}