BZOJ 1857SCOI 2010传送带
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三分套三分,虽然简单,但是也得掌握,,,
时间复杂度$O(log_{1.5}^2 n)$
一开始WA好几次发现是快速读入里没有return,这样也能过样例?
#include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const double eps = 1e-3; int in() { int k = 0, fh = 1; char c = getchar(); for(; c < ‘0‘ || c > ‘9‘; c = getchar()) if (c == ‘-‘) fh = -1; for(; c >= ‘0‘ && c <= ‘9‘; c = getchar()) k = (k << 3) + (k << 1) + c - ‘0‘; return k * fh; } double p, q, r; struct Point { double x, y; } A1, A2, B1, B2; double sqr(double x) {return x * x;} double dis(Point A, Point B) {return sqrt(sqr(A.x - B.x) + sqr(A.y - B.y));} double cal(Point B) { double c1, c2; Point P1 = B1, P2 = B2, x1, x2; while (fabs(P1.x - P2.x) > eps || fabs(P1.y - P2.y) > eps) { x1 = (Point) {P1.x + (P2.x - P1.x) / 3, P1.y + (P2.y - P1.y) / 3}; x2 = (Point) {P1.x + (P2.x - P1.x) * 2 / 3, P1.y + (P2.y - P1.y) * 2 / 3}; c1 = dis(B, x1) / r + dis(x1, B2) / q, c2 = dis(B, x2) / r + dis(x2, B2) / q; if (c1 < c2) P2 = x2; else P1 = x1; } return dis(B, P1) / r + dis(P1, B2) / q; } int main() { A1.x = in(); A1.y = in(); A2.x = in(); A2.y = in(); B1.x = in(); B1.y = in(); B2.x = in(); B2.y = in(); p = in(); q = in(); r = in(); double c1, c2; Point P1 = A1, P2 = A2, x1, x2; while (fabs(P1.x - P2.x) > eps || fabs(P1.y - P2.y) > eps) { x1 = (Point) {P1.x + (P2.x - P1.x) / 3, P1.y + (P2.y - P1.y) / 3}; x2 = (Point) {P1.x + (P2.x - P1.x) * 2 / 3, P1.y + (P2.y - P1.y) * 2 / 3}; c1 = dis(A1, x1) / p + cal(x1), c2 = dis(A1, x2) / p + cal(x2); if (c1 < c2) P2 = x2; else P1 = x1; } printf("%.2lf\n", dis(A1, P1) / p + cal(P1)); return 0; }
期末考试Bless All!
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