hdu 4033 Regular Polygon 计算几何 二分+余弦定理

题目链接

给一个n个顶点的正多边形， 给出多边形内部一个点到n个顶点的距离， 让你求出这个多边形的边长。

二分边长， 然后用余弦定理求出给出的相邻的两个边之间的夹角， 看所有的加起来是不是2Pi。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
double a[104];
int main()
{
    int t, n;
    cin>>t;
    for(int casee = 1; casee<=t; casee++) {
        printf("Case %d: ", casee);
        cin>>n;
        for(int i = 0; i<n; i++) {
            scanf("%lf", &a[i]);
        }
        double l = -inf, r = inf;
        for(int i = 0; i<n; i++) {
            l = max(l, fabs(a[i]-a[(i+1)%n]));
            r = min(r, a[i]+a[(i+1)%n]);
        }
        int flag = 0;
        while(fabs(r-l)>eps) {
            double mid = (l+r)/2, ans = 0;
            for(int i = 0; i<n; i++) {
                int tmp = (i+1)%n;
                ans += acos((a[i]*a[i]+a[tmp]*a[tmp]-mid*mid)/(2*a[i]*a[tmp]));
            }
            if(fabs(ans-2*PI)<eps) {
                flag = 1;
                break;
            }
            if(ans>2*PI)
                r = mid;
            else
                l = mid;
        }
        if(flag) {
            printf("%.3f
", l);
        } else {
            puts("impossible");
        }
    }
    return 0;
}