poj 1981(单位圆覆盖最多点问题模板)

Circle and Points

Time Limit: 5000MS		Memory Limit: 30000K
Total Submissions: 7327		Accepted: 2651
Case Time Limit: 2000MS

Description

You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enclose as many of the points as possible. Find how many points can be simultaneously enclosed at the maximum. A point is considered enclosed by a circle when it is inside or on the circle.

Fig 1. Circle and Points

Input

The input consists of a series of data sets, followed by a single line only containing a single character '0', which indicates the end of the input. Each data set begins with a line containing an integer N, which indicates the number of points in the data set. It is followed by N lines describing the coordinates of the points. Each of the N lines has two decimal fractions X and Y, describing the x- and y-coordinates of a point, respectively. They are given with five digits after the decimal point.

You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0 <= Y <= 10.0. No two points are closer than 0.0001. No two points in a data set are approximately at a distance of 2.0. More precisely, for any two points in a data set, the distance d between the two never satisfies 1.9999 <= d <= 2.0001. Finally, no three points in a data set are simultaneously very close to a single circle of radius one. More precisely, let P1, P2, and P3 be any three points in a data set, and d1, d2, and d3 the distances from an arbitrarily selected point in the xy-plane to each of them respectively. Then it never simultaneously holds that 0.9999 <= di <= 1.0001 (i = 1, 2, 3).

Output

For each data set, print a single line containing the maximum number of points in the data set that can be simultaneously enclosed by a circle of radius one. No other characters including leading and trailing spaces should be printed.

Sample Input

3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
0

Sample Output

2
5
5
11

代码转自，不想去弄了。。以后就做模板用好了 http://www.cnblogs.com/-sunshine/archive/2012/10/11/2719859.html
贴个模板：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int N = 300;
struct Point{
    double x,y;
}p[N];
struct Node{
    double angle;
    bool in;
}arc[180000];
int n,cnt;
double R;
double dist(Point p1,Point p2){
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
bool cmp(Node n1,Node n2){
    return n1.angle!=n2.angle?n1.angle<n2.angle:n1.in>n2.in;
}
void MaxCircleCover(){
    int ans=1;
    for(int i=0;i<n;i++){
        int cnt=0;
        for(int j=0;j<n;j++){
            if(i==j) continue;
            if(dist(p[i],p[j])>R*2) continue;
            double angle=atan2(p[i].y-p[j].y,p[i].x-p[j].x);
            double phi=acos(dist(p[i],p[j])/2);
            arc[cnt].angle=angle-phi;arc[cnt++].in=true;
            arc[cnt].angle=angle+phi;arc[cnt++].in=false;
        }
        sort(arc,arc+cnt,cmp);
        int tmp=1;
        for(int i=0;i<cnt;i++){
            if(arc[i].in)  tmp++;
            else tmp--;
            ans=max(ans,tmp);
        }
    }
    printf("%d
",ans);
}
int main(){
    while(scanf("%d",&n)!=EOF&&n){
        //scanf("%lf",&R);
        R = 1; //此题R为1
        for(int i=0;i<n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
            MaxCircleCover();
    }
    return 0;
}