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 < 0) 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 = 1; i <= N; i++){
            scanf("%d", &w);
            if(i == 1) st = w;
            if(i == N) ed = w;
        }
        int ans = sqrt(myabs(st-ed));
        printf("%d
", ans);
    }
    return 0;
}