POJ 1981 Circle and Points 单位圆覆盖

枚举两点,算出圆心,枚举其他点...加个优化就2MS水过了...

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define BUG         cout<<" BUG! "<<endl;
#define LINE        cout<<" ------------------ "<<endl;
#define FIN         freopen("in.txt","r",stdin);
#pragma comment     (linker,"/STACK:102400000,102400000")
template<class T> inline T L(T a)       {return (a << 1);}
template<class T> inline T R(T a)       {return (a << 1 | 1);}
template<class T> inline T lowbit(T a)  {return (a & -a);}
template<class T> inline T Mid(T a,T b) {return ((a + b) / 2);}
template<class T> inline T gcd(T a,T b) {return b ? gcd(b,a%b) : a;}
template<class T> inline T lcm(T a,T b) {return a / gcd(a,b) * b;}
template<class T> inline T Min(T a,T b) {return a < b ? a : b;}
template<class T> inline T Max(T a,T b) {return a > b ? a : b;}
template<class T> inline T Min(T a,T b,T c)     {return min(min(a,b),c);}
template<class T> inline T Max(T a,T b,T c)     {return max(max(a,b),c);}
template<class T> inline T Min(T a,T b,T c,T d) {return min(min(a,b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d) {return max(max(a,b),max(c,d));}
template<class T> inline void mem(T &a,int b)   {memset(a,b,sizeof(a));}
template<class T> inline T exGCD(T a, T b, T &x, T &y){ //
    if(!b) return x = 1, y = 0, a;
    T res = exGCD(b, a % b, x, y), tmp = x;
    x = y, y = tmp - (a / b) * y; return res;
}
typedef long long LL;    typedef unsigned long long ULL;
//typedef __int64 LL;      typedef unsigned __int64 ULL;
const LL LINF   = 1LL << 60;
const int MOD   = 1000000007;
const int INF   = 0x7fffffff;
const int MAXN  = 300+5;//100000+5;
const double EPS    = 1e-8;
const double DINF   = 1e15;
const double PI     = acos(-1.0);

/*********************   By  F   *********************/
struct POINT {
    double x,y;
    POINT(double _x = 0,double _y = 0){x = _x, y = _y;}
}p[MAXN];
struct CIRCLE {
    double r;
    POINT o;
    CIRCLE(double _r = 0 , POINT _o = 0){r = _r, o = _o;}
};
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 incir(POINT a ,CIRCLE b){
    if((a.x - b.o.x) * (a.x - b.o.x) + (a.y - b.o.y) * (a.y - b.o.y) - 1 <= EPS) return true;
    return false;
}
POINT get_p(POINT p1,POINT p2){
    POINT tmp = POINT(Mid(p1.x,p2.x),Mid(p1.y,p2.y));
    double angle = atan2(p1.x - p2.x,p2.y - p1.y);
    double dis = sqrt(1 - dist(p1,tmp) * dist(p1,tmp));
    return POINT(tmp.x + dis * cos(angle),tmp.y + dis * sin(angle));
}
int n ;
int main(){
    //FIN
    while(scanf("%d",&n)!=EOF && n){
        for(int i = 0 ; i < n ; i++)
           scanf("%lf%lf",&p[i].x,&p[i].y);
        int cnt = 1;
        for(int i = 0 ; i < n ; i++){
            for(int j = i + 1 ; j < n ; j++){
                if(dist(p[i],p[j]) > 2.0) continue;
                CIRCLE c = CIRCLE(1.0,get_p(p[i],p[j]));
                int cc = 0;
                for(int k = 0 ; k < n ; k++)
                    if(incir(p[k],c)) cc++;
                cnt = Max(cnt,cc);
            }
        }
        printf("%d
",cnt);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/Felix-F/p/3303120.html