HDU - 6343 Graph Theory Homework 【水题】

原创于 2019-06-02 16:39:15 发布 · 322 阅读

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博客是一道图论作业题，给定一个含n个顶点的完全图，各顶点有权值wi，两点间距离为⌊∣wi−wj∣⌋，要求计算从顶点1到n的最短路径长度。通过均值与绝对值三角不等式分析得出，任意两点直接连接最短。

Graph Theory Homework

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题目

There is a complete graph containing $n$ vertices, the weight of the $i$ -th vertex is $w_i$ .
The length of edge between vertex $i$ and $j$ $\ne j)$ is $\lfloor \sqrt{|w_i - w_j|} \rfloor$ .
Calculate the length of the shortest path from $1$ to $n$ .

Input:

The first line of the input contains an integer $T$ $\le T \le 10)$ denoting the number of test cases.
Each test case starts with an integer $n$ $\le n \le 10 ^ 5)$ denoting the number of vertices in the graph.
The second line contains $n$ integers, the $i$ -th integer denotes $w_i$ $\le w_i \le 10 ^ 5)$ .

Output:

For each test case, print an integer denoting the length of the shortest path from $1$ to $n$ .

题目大意

给你一个完全图，然后每个节点都有一个权值 $w_i$ ，每两个点之间的距离 $\lfloor \sqrt{|w_i - w_j|} \rfloor$ ，求从1到n的最短距离是多少

分析

由均值不等式与绝对值三角不等式有：
$\sqrt{|a - b|} + \sqrt{|b - c|} \ge \sqrt{|a - b| + |b - c|} \ge \sqrt{|a - c|}$
显然，任意两点之间，直接连接是最短的。

参考代码

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#ifdef LOCAL
#define frein(x) freopen(x, "r", stdin)
#define freout(x) freopen(x, "w", stdout)
#else
#define frein(x)
#define freout(x)
#endif
#define mem(x, v)  memset(x, v, sizeof(x))
#define squ(x) ((x) * (x))
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 7;
const int inf = 0x3f3f3f3f;
int arr[maxn];
int main(void){
    //frein("data.in");
    //freout("data.out");
    int T; cin >> T;
    while(T--){
        int n; cin >> n;
        for(int i = 1; i <= n; i++){
            cin >> arr[i];
        }
        int ans = sqrt(abs(arr[1] - arr[n]));
        cout << ans << endl;
    }
    return 0;
}