Codeforces 101173 C - Convex Contour

Posted 2020-12-31 widsom

思路：

如果所有的图形都是三角形，那么答案是2*n+1

否则轮廓肯定触到了最上面，要使轮廓线最短，那么轮廓肯定是中间一段平的

我们考虑先将轮廓线赋为2*n+2，然后删去左右两边多余的部分

如果最左边或最由边是正方形，那么不需要删

如果最左边或最由边是圆形，那么删取2 - pi/2

如果如果最左边或最由边是三角形，那么我们需要找到一段连续的三角形，然后考虑怎么删去

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

int main() {
    fio;
    int n;
    string s;
    cin >> n;
    cin >> s;
    bool f = true;
    for (int i = 0; i < n; i++) if(s[i] != ‘T‘) f = false;
    if(f) {
        cout << 2*n + 1 << endl;
    }
    else {
        double ans = 2*n + 2;
        if(s[0] == ‘T‘) {
            int x = 1;
            for (int i = 1; i < n; i++) {
                if(s[i] == ‘T‘) {
                    x++;
                }
                else if(s[i] == ‘C‘){
                    double tot = x + 0.5;
                    double t = sqrt((sqrt(3)/2 - 0.5)*(sqrt(3)/2 - 0.5) + x*x - 0.25);
                    tot -= t;
                    double a = atan((sqrt(3)/2 - 0.5) / x);
                    double b = atan(t*2);
                    double c = pi/2 -a - b;
                    tot -= c*0.5;
                    ans -= tot;
                    break;
                }
                else if(s[i] == ‘S‘) {
                    double tot = x;
                    double t = sqrt((1-sqrt(3)/2)*(1-sqrt(3)/2) + (x-0.5) * (x-0.5));
                    tot -= t;
                    ans -= tot;
                    break;
                }
            }
        }
        else if(s[0] == ‘C‘) {
            ans -= 2 - pi/2;
        }

        reverse(s.begin(), s.end());
        if(s[0] == ‘T‘) {
            int x = 1;
            for (int i = 1; i < n; i++) {
                if(s[i] == ‘T‘) {
                    x++;
                }
                else if(s[i] == ‘C‘){
                    double tot = x + 0.5;
                    double t = sqrt((sqrt(3)/2 - 0.5)*(sqrt(3)/2 - 0.5) + x*x - 0.25);
                    tot -= t;
                    double a = atan((sqrt(3)/2 - 0.5) / x);
                    double b = atan(t*2);
                    double c = pi/2 -a - b;
                    tot -= c*0.5;
                    ans -= tot;
                    break;
                }
                else if(s[i] == ‘S‘) {
                    double tot = x;
                    double t = sqrt((1-sqrt(3)/2)*(1-sqrt(3)/2) + (x-0.5) * (x-0.5));
                    tot -= t;
                    ans -= tot;
                    break;
                }
            }
        }
        else if(s[0] == ‘C‘) {
            ans -= 2 - pi/2;
        }
        cout << fixed << setprecision(9) << ans << endl;
    }
    return 0;
}