2018 Multi-University Training Contest 4 Problem L. Graph Theory Homework 【YY】

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：

 #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 ;
 }