2018 Multi-University Training Contest 4.Problem L. Graph Theory Homework（简单数学）

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这道题目是2018年多校训练赛4题L，涉及图论问题。给定一个包含n个顶点的完全图，每个顶点有特定权重，任意两个顶点间边的长度为两顶点权重差的平方根向下取整。求从1到n的最短路径长度。输入包含测试用例数T和每个测试用例的顶点数n及各顶点权重，输出最短路径长度。样例输入为1个测试用例，3个顶点，权重分别为1、3、5，输出结果为2。

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Problem L. Graph Theory Homework

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 378 Accepted Submission(s): 235

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

Source

2018 Multi-University Training Contest 4

解题思路：两点之间的直接距离比经过第三点中转的距离小。

#include <bits/stdc++.h>
using namespace std;

void solve () {
    int n, a; cin >> n;
    for (int i = 1; i <= n; ++i) {
        int x; cin >> x;
        if (i == 1) a = x;
        if (i == n) cout << (int)sqrt(fabs(x - a) + 1e-3) << endl;
    }
}

int main () {
    ios::sync_with_stdio(false);
    int T; cin >> T;
    while (T--) solve();
    return 0;
}