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。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>

using namespace std;

#define LL long long
#define eps 1e-10
#define mxn 1000200
#define mxe 2000020
#define inf 1e12
#define Pi acos( -1.0 )

//精度
int dcmp( double x ){
    if( fabs( x ) < eps ) return 0;
    return x < 0 ? -1 : 1;
}
// 点
struct point{
    double x, y;
    point( double x = 0, double y = 0 ) : x(x), y(y) {}
    point operator + ( const point &b ) const{
        return point( x + b.x, y + b.y );
    }
    point operator - ( const point &b ) const{
        return point( x - b.x, y - b.y );
    }
    point operator * ( const double &k ) const{
        return point( x * k, y * k );
    }
    point operator / ( const double &k ) const{
        return point( x / k, y / k );
    }
    bool operator < ( const point &b ) const{
        return dcmp( x - b.x ) < 0 || dcmp( x - b.x ) == 0 && dcmp( y - b.y ) < 0;
    }
    bool operator == ( const point &b ) const{
        return dcmp( x - b.x ) == 0 && dcmp( y - b.y ) == 0;
    }
    double len(){
        return sqrt( x * x + y * y );
    }
};
typedef point Vector;
// 点积
double dot( Vector a, Vector b ){
    return a.x * b.x + a.y * b.y;
}
// 叉积
double cross( Vector a, Vector b ){
    return a.x * b.y - a.y * b.x;
}
//两圆相交
void circle_cross( point a, double ra, point b, double rb, point &v1, point &v2 ){
    double d = ( a - b ).len();
    double da = ( ra * ra + d * d - rb * rb ) / ( 2 * ra * d );
    double aa = atan2( (b-a).y, (b-a).x );
    double rad = acos( da );
    v1 = point( a.x + cos( aa - rad ) * ra, a.y + sin( aa - rad ) * ra );
    v2 = point( a.x + cos( aa + rad ) * ra, a.y + sin( aa + rad ) * ra );
}
double R;
point s0, s1, s2;
void solve(){
    point ans1, ans2;
    if( dcmp( (s0-s1).len() - 2 * R ) <= 0 && dcmp( (s0-s2).len() - 2 * R ) <= 0 ){
        ans1 = (s0+s1)/2, ans2 = (s0+s2)/2;
    }
    else if(dcmp( (s0-s1).len() - 2 * R ) <= 0 || dcmp( (s0-s2).len() - 2 * R ) <= 0){
        if( dcmp( (s0-s2).len() - 2 * R ) <= 0 ) swap( s1, s2 );
        point v1, v2;
        double tmp;
        circle_cross( s0, R, s1, R, v1, v2 ); //v1,v2扇形交点
        tmp = (s2-v1).len(), ans1 = v1, ans2 = (s2+v1)/2;
        if( (s2-v2).len() < tmp )
            tmp = (s2-v2).len(), ans1 = v2, ans2 = (s2+v2)/2;
        double len1 = (s2-s0).len(), len2 = (s2-s1).len();
        point u1 = s0 + ( s2 - s0 ) * (R/len1);
        if( (u1-s1).len() <= R ){
            if( (s2-u1).len() < tmp )
                tmp = (s2-u1).len(), ans1 = u1, ans2 = (s2+u1)/2;
        }
        point u2 = s1 + ( s2 - s1 ) * (R/len2);
        if( (u2-s0).len() <= R ){
            if( (s2-u2).len() < tmp )
                tmp = (s2-u2).len(), ans1 = u2, ans2 = (s2+u2)/2;
        }
        if( tmp > 2 * R ){
            puts( "Death" ); return ;
        }
    }
    else{
        if( dcmp( (s1-s2).len() - 2 * R ) > 0 ){
            puts( "Death" );return ;
        }
        point v1, v2;
        circle_cross( s1, R, s2, R, v1, v2 );
        double tmp;
        tmp = (s0-v1).len(), ans1 = v1, ans2 = (s0+v1)/2;
        if( (s0-v2).len() < tmp )
            tmp = (s0-v2).len(), ans1 = v2, ans2 = (s0+v2)/2;
        double len1 = (s0-s1).len(), len2 = (s0-s2).len();
        point u1 = s1 + (s0-s1) * (R/len1), u2 = s2 + (s0-s2)* (R/len2);
        if( (u1-s2).len() <= R ){
            if( (s0-u1).len() < tmp )
                tmp = (s0-u1).len(), ans1 = u1, ans2 = (u1+s0)/2;
        }
        if( (u2-s1).len() <= R ){
            if( (s0-u2).len() < tmp ){
                tmp = (s0-u2).len(), ans1 = u2, ans2 = (u2+s0)/2;
            }
        }
        if( tmp > 2 * R ){
            puts( "Death" ); return;
        }
    }
    puts( "Now we have enough power");
    printf( "%.5lf %.5lf
%.5lf %.5lf
", ans1.x, ans1.y, ans2.x, ans2.y );
}
int main(){
    scanf( "%lf", &R );
    scanf( "%lf %lf", &s0.x, &s0.y );
    scanf( "%lf %lf", &s1.x, &s1.y );
    scanf( "%lf %lf", &s2.x, &s2.y );
    solve();
    return 0;
}