传送门：http://acm.hdu.edu.cn/showproblem.php?pid=6343
Problem L. Graph Theory Homework
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1536 Accepted Submission(s): 830
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.
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).
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
Source
题意概括：
给出每个点的权值，点与点之间的距离等于 √| wi-wj | ，求起点到终点的最短距离。
解题思路：
一道伪装成图论的水题。
最短距离就是两点距离，因为如果中间放入其他点来进行更新路径是不会获得更短的路径的。
因为 √a + √b > √(a+b) ;
AC code：
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
#include<cmath>
#include<set>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
int N; int myabs(int x)
{
if(x < ) return -x;
return x;
} int main()
{
int w, st, ed;
int T_case;
scanf("%d", &T_case);
while(T_case--){
scanf("%d", &N);
for(int i = ; i <= N; i++){
scanf("%d", &w);
if(i == ) st = w;
if(i == N) ed = w;
}
int ans = sqrt(myabs(st-ed));
printf("%d\n", ans);
}
return ;
}