Max Angle //2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

Max Angle

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 12   Accepted Submission(s) : 5

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Problem Description

Given many points in a plane, two players are playing an interesting game.

Player1 selects one point A as the vertex of an angle. Then player2 selects other two points B and C. A, B and C are different with each other. Now they get an angle B-A-C.

Player1 wants to make the angle as large as possible, while player2 wants to make the angle as small as possible.

Now you are supposed to find the max angle player1 can get, assuming play2 is c lever enough.

Input

There are many test cases. In each test case, the first line is an integer n (3 <= n <= 1001), which is the number of points on the plane. Then there are n lines. Each contains two floating number x, y, witch is the coordinate of one point. n <= 0 denotes the end of input.

Output

For each test case, output just one line, containing the max angle player1 can get in degree format. The result should be accurated up to 4 demicals.

Sample Input

3
0 0
2 0
0 5
-1

Sample Output

90.0000

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double pi=acos(-1+0.0);
struct node
{
    double x,y;
};
double dis(node a, node b)
{
     return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}
int main()
{
    int n;
    node a[1100];
    double t[1100];
    while(scanf("%d",&n)==1&&n>0)
    {
        for(int i=0;i<n;i++) scanf("%lf%lf",&a[i].x,&a[i].y);
        double cnt=0;
        for(int i=0;i<n;i++)
        {
            int l=0;
            for(int j=0;j<n;j++)
            {
                if(i==j) continue;
                t[l++]=acos((a[j].x-a[i].x)/dis(a[i],a[j]));
                if(a[j].y<a[i].y) t[l-1]=2*pi-t[l-1];
            }
            sort(t,t+l);
            double _min=2*pi-t[l-1]+t[0];
            for(int i=1;i<l;i++) if(t[i]-t[i-1]<_min) _min=t[i]-t[i-1];
            if(_min>cnt) cnt=_min;
        }
        printf("%.4lf/n",cnt*180/pi);
    }
    return 0;
}