URAL 1913 Titan Ruins: Old Generators Are Fine Too
不知道怎么就1A了。。题目意思不难理解，设水晶的坐标为s0，两个发动机的坐标是s1，s2，半径为R，就分三种情况。。第一种情况就是s1，s2到s0的距离都小于2*R，这种情况，两个人的位置就是s1-s0，s2-s0的中点；第二种情况就是s1和s2只有一个到s0的距离小于2*R，假设s1相交，这个时候s1与s0相交的交点，两个交点所在的圆弧里面就是其中一个人的坐标，另外一个人的坐标有可能在两个交点上，也有可能在圆弧里面；第三种情况就是s1和s2到s0的距离大于2*R，但是s1和s2的距离小于2*R，这个时候其中一个人的坐标就是在s1和s2交点所在的圆弧里面，另一个人的坐标根据s0到s1，s2还有圆弧交点来判断；最后一种情况就是以上三种都不满足，就输出death。。具体还是看代码，然后画一画，就明白了。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <map> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-10 12 #define mxn 1000200 13 #define mxe 2000020 14 #define inf 1e12 15 #define Pi acos( -1.0 ) 16 17 //精度 18 int dcmp( double x ){ 19 if( fabs( x ) < eps ) return 0; 20 return x < 0 ? -1 : 1; 21 } 22 // 点 23 struct point{ 24 double x, y; 25 point( double x = 0, double y = 0 ) : x(x), y(y) {} 26 point operator + ( const point &b ) const{ 27 return point( x + b.x, y + b.y ); 28 } 29 point operator - ( const point &b ) const{ 30 return point( x - b.x, y - b.y ); 31 } 32 point operator * ( const double &k ) const{ 33 return point( x * k, y * k ); 34 } 35 point operator / ( const double &k ) const{ 36 return point( x / k, y / k ); 37 } 38 bool operator < ( const point &b ) const{ 39 return dcmp( x - b.x ) < 0 || dcmp( x - b.x ) == 0 && dcmp( y - b.y ) < 0; 40 } 41 bool operator == ( const point &b ) const{ 42 return dcmp( x - b.x ) == 0 && dcmp( y - b.y ) == 0; 43 } 44 double len(){ 45 return sqrt( x * x + y * y ); 46 } 47 }; 48 typedef point Vector; 49 // 点积 50 double dot( Vector a, Vector b ){ 51 return a.x * b.x + a.y * b.y; 52 } 53 // 叉积 54 double cross( Vector a, Vector b ){ 55 return a.x * b.y - a.y * b.x; 56 } 57 //两圆相交 58 void circle_cross( point a, double ra, point b, double rb, point &v1, point &v2 ){ 59 double d = ( a - b ).len(); 60 double da = ( ra * ra + d * d - rb * rb ) / ( 2 * ra * d ); 61 double aa = atan2( (b-a).y, (b-a).x ); 62 double rad = acos( da ); 63 v1 = point( a.x + cos( aa - rad ) * ra, a.y + sin( aa - rad ) * ra ); 64 v2 = point( a.x + cos( aa + rad ) * ra, a.y + sin( aa + rad ) * ra ); 65 } 66 double R; 67 point s0, s1, s2; 68 void solve(){ 69 point ans1, ans2; 70 if( dcmp( (s0-s1).len() - 2 * R ) <= 0 && dcmp( (s0-s2).len() - 2 * R ) <= 0 ){ 71 ans1 = (s0+s1)/2, ans2 = (s0+s2)/2; 72 } 73 else if(dcmp( (s0-s1).len() - 2 * R ) <= 0 || dcmp( (s0-s2).len() - 2 * R ) <= 0){ 74 if( dcmp( (s0-s2).len() - 2 * R ) <= 0 ) swap( s1, s2 ); 75 point v1, v2; 76 double tmp; 77 circle_cross( s0, R, s1, R, v1, v2 ); //v1,v2扇形交点 78 tmp = (s2-v1).len(), ans1 = v1, ans2 = (s2+v1)/2; 79 if( (s2-v2).len() < tmp ) 80 tmp = (s2-v2).len(), ans1 = v2, ans2 = (s2+v2)/2; 81 double len1 = (s2-s0).len(), len2 = (s2-s1).len(); 82 point u1 = s0 + ( s2 - s0 ) * (R/len1); 83 if( (u1-s1).len() <= R ){ 84 if( (s2-u1).len() < tmp ) 85 tmp = (s2-u1).len(), ans1 = u1, ans2 = (s2+u1)/2; 86 } 87 point u2 = s1 + ( s2 - s1 ) * (R/len2); 88 if( (u2-s0).len() <= R ){ 89 if( (s2-u2).len() < tmp ) 90 tmp = (s2-u2).len(), ans1 = u2, ans2 = (s2+u2)/2; 91 } 92 if( tmp > 2 * R ){ 93 puts( "Death" ); return ; 94 } 95 } 96 else{ 97 if( dcmp( (s1-s2).len() - 2 * R ) > 0 ){ 98 puts( "Death" );return ; 99 } 100 point v1, v2; 101 circle_cross( s1, R, s2, R, v1, v2 ); 102 double tmp; 103 tmp = (s0-v1).len(), ans1 = v1, ans2 = (s0+v1)/2; 104 if( (s0-v2).len() < tmp ) 105 tmp = (s0-v2).len(), ans1 = v2, ans2 = (s0+v2)/2; 106 double len1 = (s0-s1).len(), len2 = (s0-s2).len(); 107 point u1 = s1 + (s0-s1) * (R/len1), u2 = s2 + (s0-s2)* (R/len2); 108 if( (u1-s2).len() <= R ){ 109 if( (s0-u1).len() < tmp ) 110 tmp = (s0-u1).len(), ans1 = u1, ans2 = (u1+s0)/2; 111 } 112 if( (u2-s1).len() <= R ){ 113 if( (s0-u2).len() < tmp ){ 114 tmp = (s0-u2).len(), ans1 = u2, ans2 = (u2+s0)/2; 115 } 116 } 117 if( tmp > 2 * R ){ 118 puts( "Death" ); return; 119 } 120 } 121 puts( "Now we have enough power"); 122 printf( "%.5lf %.5lf\n%.5lf %.5lf\n", ans1.x, ans1.y, ans2.x, ans2.y ); 123 } 124 int main(){ 125 scanf( "%lf", &R ); 126 scanf( "%lf %lf", &s0.x, &s0.y ); 127 scanf( "%lf %lf", &s1.x, &s1.y ); 128 scanf( "%lf %lf", &s2.x, &s2.y ); 129 solve(); 130 return 0; 131 }
