2021ICPC上海站热身赛

链接：牛客竞赛_ACM/NOI/CSP/CCPC/ICPC算法编程高难度练习赛_牛客竞赛OJ

A题签到题：

#include 
#define x first
#define y second
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
int y[N];
double cut[N];
double dis(int a, int b) // a, b两点距离
{
    int x = b - a;
    return sqrt(x * x + (y[a] - y[b]) * (y[a] - y[b]));
}
int main()
{
    // freopen("C:/Users/zhaochen/Desktop/input.txt", "r", stdin);
    cin.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> y[i];
    }
    double sum = 0;
    double maxcut = 0;
    int del = 0;
    for (int i = 1; i <= n; i++)
    {
        sum += dis(i, i - 1);
        cut[i] = dis(i, i - 1) + dis(i + 1, i) - dis(i - 1, i + 1); // 删去这个点能减少的长度
        if (cut[i] > maxcut)
        {
            del = i;
            maxcut = cut[i];
        }
    }
    sum += dis(n, n+1);
    sum -= maxcut;
    cut[del] = 0; // 删去点del后，更新他前后的两个点
    cut[del - 1] = dis(del - 1, del - 2) + dis(del + 1, del - 1) - dis(del - 2, del + 1);
    cut[del + 1] = dis(del + 2, del + 1) + dis(del + 1, del - 1) - dis(del - 1, del + 2);
    maxcut = 0;
    for (int i = 1; i <= n; i++)
    {
        maxcut = max(cut[i], maxcut);
    }
    sum -= maxcut;
    printf("%.16f", sum);
    return 0;
}