UVA 11355 Cool Points( 極角計算 )

標籤：

We have a circle of radius R and several line segments situated within the circumference of this circle. Let’s define a cool point to be a point on the circumference of this circle so that the line segment that is formed by this point and the centre of the circle makes no intersection with any of the given line segments.

For this problem, you have to find out the percentage of cool points from all possible points on the circumference of the given circle.

Input

The input file starts with an integer T(T<1000) that indicates the number of test cases. Each case starts with 2 integers N(0 <= N < 100) and R(0 < R < 1001). N represents the number of line segments and R represents the radius of the circle. Each of the next N lines contains 4 integers in the order x1, y1, x2 and y2. (x1, y1) – (x2, y2) represents a line segment.

You can assume that all the line segments will be inside the circle and no line segment passes through the origin. Also consider the center of the circle to be on the origin.

Output

For each input, output the case number followed by the percentage, rounded to 2 decimal places, of cool points. Look at the output for exact format.

 

Sample Input	Output for Sample Input
2 1 10 2 0 0 2 0 5	Case 1: 75.00% Case 2: 100.00%

 

 

 

題意就是給出一個圓 ， 還有一些線段（ 不過圓心 ）

問在圓上的點與圓心的連線沒有交點的點佔總點數的百分比。

那麼就將所有線段的兩個端點都弄成極角。然後排個序。

如果線段的兩個端點極角分別跨越正負的話，就將它處理成兩條線段 。 

再處理一下區間 ， 求百分比就能過了。

 

#include<bits/stdc++.h>using namespace std;const int N = 205;const double PI = acos(-1.0);const double eps = 1e-6;typedef pair<double,double> pii;#define X first#define Y secondint n , top ;vector<pii>p;double x[N] , y[N];int dcmp( double x ) {    if( fabs(x) < eps ) return 0 ;    return x<0?-1:1;}double cal( double avg1 , double avg2 ) {    if( avg1 < 0 && avg2 < 0 ) return -1.0 ;    if( avg1 > 0 && avg2 > 0 ) return -1.0 ;    return fabs( avg1 ) + fabs( avg2 ) ;}void Solve() {    top = 0 ;    for( int i = 0 ; i < p.size() ; ++i ){        while( top > 0 && dcmp( p[i].X-x[top-1])==0 && dcmp(p[i].Y-y[top-1])>=0 )top--;        if( top > 0 && dcmp(x[top-1]-p[i].X)<=0 && dcmp(y[top-1]-p[i].Y)>=0 )continue ;        x[top] = p[i].X , y[top] = p[i].Y , top++;    }}void Run() {    double x1 , y1 , x2 , y2 , r ;    scanf("%d%lf",&n,&r); p.clear();    if( !n ) { puts("100.00%"); return ; }    for( int i = 0 ; i < n ; ++i ){        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);        double avg1 = atan2(y1,x1) , avg2 = atan2(y2,x2) ;        if( avg1 > avg2 ) swap( avg1 , avg2 );        double d = cal( avg1 , avg2 );        if( d == -1.0 ) { p.push_back(pii(avg1,avg2)); continue; }        if( d < PI ) {            p.push_back(pii(avg1,0));            p.push_back(pii(0,avg2));        }        else {            p.push_back(pii(-PI,avg1));            p.push_back(pii(avg2,PI));        }    }    sort( p.begin() , p.end() );    Solve();    double s = x[0] , e = y[0] , fenzi = 0 ;    for( int i = 1 ; i < top ; ++i ){        if( x[i] > e ) {            fenzi += ( e - s ); s = x[i];        }        e = y[i] ;    }    fenzi += ( e - s );    double ans = 100.0 - fenzi/(2.0*PI)*100.0;    printf("%.2lf",ans);puts("%");}int main() {//    freopen("in.txt","r",stdin);    int _ , cas = 1 ;    scanf("%d",&_); while( _ -- ) {        printf("Case %d: ",cas++);        Run();    }}

View Code

 

UVA 11355 Cool Points( 極角計算 )