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.

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

Source

2018 Multi-University Training Contest 4

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：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<vector>
 6 #include<queue>
 7 #include<cmath>
 8 #include<set>
 9 #define INF 0x3f3f3f3f
10 #define LL long long
11 using namespace std;
12 int N;
13
14 int myabs(int x)
15 {
16     if(x < 0) return -x;
17     return x;
18 }
19
20 int main()
21 {
22     int w, st, ed;
23     int T_case;
24     scanf("%d", &T_case);
25     while(T_case--){
26         scanf("%d", &N);
27         for(int i = 1; i <= N; i++){
28             scanf("%d", &w);
29             if(i == 1) st = w;
30             if(i == N) ed = w;
31         }
32         int ans = sqrt(myabs(st-ed));
33         printf("%d\n", ans);
34     }
35     return 0;
36 }

原文地址：https://www.cnblogs.com/ymzjj/p/10330782.html