题意:给一个圆C和圆心O,P、Q是圆上或圆内到圆心距离相等的两个点,在圆上取一点D,求|PD| + |QD|的最小值
析:首先这个题是可以用三分过的,不过也太,。。。。
官方题解:
很不幸不总是中垂线上的点取到最小值,考虑点在圆上的极端情况。
做P点关于圆的反演点P‘,OPD与ODP‘相似,相似比是|OP| : r。
Q点同理。
极小化PD+QD可以转化为极小化P‘D+Q‘D。
当P‘Q‘与圆有交点时,答案为两点距离,否则最优值在中垂线上取到。
时间复杂度 O(1)O(1)
也有代数做法,结论相同。
优秀的黄金分割三分应该也是可以卡过的。
主要是可以先通过反演来获得P‘,Q‘,坐标,可以通过人为再构造一个横纵坐标的比例等于OP和OP‘的比来计算,然后再考虑是不是与圆相交,这个可以用求中点,带入圆的方程得到,如果在外面,先可以求出P,Q的中点E,然后能求OE,PE,就可以求得DE,最后再求那PP‘即可。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 10000 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } double dist(double x1, double y1, double x2 = 0.0, double y2 = 0.0){ return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)); } int main(){ int T; cin >> T; while(T--){ double xp, yp, xq, yq, r; scanf("%lf", &r); scanf("%lf %lf %lf %lf", &xp, &yp, &xq, &yq); double op = dist(xp, yp); double oq = dist(xq, yq); double opp = r * r / op; double oqq = r * r / oq; double xpp = opp * xp / op; double xqq = oqq * xq / oq; double ypp = opp * yp / op; double yqq = oqq * yq / oq; double avx = (xpp + xqq) / 2.0; double avy = (ypp + yqq) / 2.0; if(avx * avx + avy * avy <= r * r){ printf("%.10f\n", dist(xpp, ypp, xqq, yqq) * op / r); continue; } avx = (xp + xq) / 2.0; avy = (yp + yq) / 2.0; double oav = r - dist(avx, avy); double tmp = dist(xp, yp, avx, avy); double ans = sqrt(tmp*tmp + oav*oav); printf("%.10f\n", ans*2.0); } return 0; }
时间: 2024-10-09 23:14:02