SOJ2658.LineDist

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2658. LineDist

	Time Limit: 1sec    Memory Limit:256MB Description In n-dimension space, a line can be determined by two distinct points. Your task is to calculate the distance between the two lines. Distance between two lines is defined as the minimal distance of two points in the two lines respectively. Distance of two points A(x1,x2,…xn) and B(y1,y2,…,yn) is defined as sqrt( (x1-y1)^2 + (x2-y2)^2 + … + (xn-yn)^2 ). Input There are several test cases. For each case, there are five lines, the first line contains the integer number n, and then each of the next four lines contains n integer numbers, these data are for four points A, B, C and D. A and B are distinct and determines line AB, C and D are distinct and determines line CD, you should calculate distance between AB and CD. Constraints: 2 <= n <= 3, other input numbers will be between -1000 and 1000. Input is ended with n = 0. Output Output each result in a single line, with two digit precision after the decimal point. Sample Input 2 0 0 0 1 1 0 1 1 2 2 0 0 2 -1 0 0 1 3 0 0 0 0 0 1 1 1 0 1 1 1 3 0 0 0 0 0 1 0 1 0 1 1 0 0 Sample Output 1.00 0.00 1.41 1.00 題意： 給兩條2維或者3維的直線求最短距離 大致思路： 對與一點，一條直線上的點與這個固定點的距離是個單峯函數；若這個點在某條線上，那麼這個點與另外一條直線的最短距離也是個單峯函數，所以 用兩重三分。 由於座標在[-1000, 1000],所以最短距離所在點的絕對值會在[0, 10003]內。 程序： #include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define MAXN #define eps (1e-13) #define INF 1100000000.0 #define abs(x) ( (x) > 0? (x): -(x) ) #define sqr(x) ((x) * (x)) #define MAX(a, b) ((a) > (b)? (a): (b)) #define MIN(a, b) ((a) < (b)? (a): (b)) typedef long long LL; struct POINT { double x, y, z; POINT( double x = 0, double y = 0, double z = 0 ): x(x), y(y), z(z) {} POINT operator - ( const POINT&d ) { return POINT( x - d.x, y - d.y, z - d.z ); } POINT operator + ( const POINT &d ) { return POINT( x + d.x, y + d.y, z + d.z ); } POINT operator * ( double d ) { return POINT( x * d, y * d, z * d ); } }; void swap( int &x, int &y ) { int temp = x; x = y; y = temp; } double dis( POINT p, POINT q ) { return sqrt( sqr( p.x - q.x ) + sqr( p.y - q.y ) + sqr( p.z - q.z ) ); } double check( POINT u, POINT x, POINT y ) { double p = 0, q = 1; POINT z = y - x; while ( abs( q - p ) > eps ) { double l = p + ( q - p ) / 3; double r = p + ( q - p ) / 3 * 2; if ( dis( u, x + z * l ) < dis( u, x + z * r ) ) q = r; else p = l; } return dis( u, x + z * p ); } void repair( POINT &p, POINT &q ) { POINT E = p - q; double u, v; if ( abs( E.x ) < eps ) { if ( abs( E.y ) < eps ) { u = -( p.z + INF ) / E.z; v = -( q.z - INF ) / E.z; } else { u = -( p.y + INF ) / E.y; v = -( q.y - INF ) / E.y; } } else { u = -( p.x + INF ) / E.x; v = -( q.x - INF ) / E.x; } p = p + E * u; q = q + E * v; } int main() { int n; while ( scanf( "%d", &n ) != EOF && n ) { POINT A, B, C, D, E; if ( n == 2 ) { scanf( "%lf%lf", &A.x, &A.y ); A.z = 0; scanf( "%lf%lf", &B.x, &B.y ); B.z = 0; scanf( "%lf%lf", &C.x, &C.y ); C.z = 0; scanf( "%lf%lf", &D.x, &D.y ); D.z = 0; } else { scanf( "%lf%lf%lf", &A.x, &A.y, &A.z ); scanf( "%lf%lf%lf", &B.x, &B.y, &B.z ); scanf( "%lf%lf%lf", &C.x, &C.y, &C.z ); scanf( "%lf%lf%lf", &D.x, &D.y, &D.z ); } repair( A, B ); repair( C, D ); E = B - A; double p = 0, q = 1; while ( abs( p - q ) > eps ) { double l = p + ( q - p ) / 3; double r = p + ( q - p ) / 3 * 2; if ( check( A + E * l, C, D ) < check( A + E * r, C, D ) ) q = r; else p = l; } double ans = check( A + E * p, C, D ); printf( "%.2lf/n", ans ); } return 0; }