Fruit Slicer

题目链接

切水果几何题，精度感觉特别坑，转化为long long 来做。具体代码如下。需要思考思考啊。

 

#include <bits/stdc++.h>

#define rep(i , a , b) for(int i=(a);i<=(b);i++)
using namespace std;
typedef long long ll;

struct Point {
    ll x , y;

    Point () {}

    Point (double _x , double _y) {
        x = _x;
        y = _y;
    }

    Point operator- (const Point &b) const {
        return Point (x - b.x , y - b.y);
    }

    ll operator^ (const Point &b) const {
        return x * b.y - y * b.x;
    }

    ll operator* (const Point &b) const {
        return x * b.x + y * b.y;
    }
};


struct Circle {
    Point c;
    ll r;

    Circle () {}

    Circle (Point c , ll r) : c (c) , r (r) {}

    bool operator< (const Circle &a) const {
        if ( a.c.x != c.x ) return c.x < a.c.x;
        return c.y < a.c.y;
    }
} e[111];

ll dist (Point p1 , ll a , ll b , ll c) {
    ll x = p1.x;
    ll y = p1.y;
    if ( abs (x * a + y * b + c) >= 1e9 ) return - 1;
    return ( x * a + y * b + c ) * ( x * a + y * b + c );
}

bool inter (ll a , ll b , ll c , Circle c1) {
    if ( dist (c1.c , a , b , c) == - 1 ) return false;
    return dist (c1.c , a , b , c) <= 1ll * 40000 * ( a * a + b * b );
}

bool same (Circle a , Circle b) {
    if ( a.c.x == b.c.x && a.c.y == b.c.y ) return true;
    return false;
}

ll read (string s) {
    ll op;
    ll ans = 0;
    if ( s[ 0 ] == '-' ) op = - 1;
    else op = 1;
    for ( int i = 0 ; i < s.length () ; i ++ ) {
        if ( s[ i ] == '.' || s[ i ] == '-' ) continue;
        ans = ans * 10 + s[ i ] - '0';
    }
    return ans * op;
}

int main () {
    int n;
    scanf ("%d" , &n);
    for ( int i = 1 ; i <= n ; i ++ ) {
        string s1 , s2;
        cin >> s1 >> s2;
        e[ i ].c.x = read (s1);
        e[ i ].c.y = read (s2);
        e[ i ].r = 1ll * 100;
    }
    if ( n == 1 ) {
        printf ("1\n");
        return 0;
    }
    if ( n == 2 ) {
        printf ("2\n");
        return 0;
    }
    int ans = 0;
    sort (e + 1 , e + 1 + n);
    rep (i , 1 , n) {
        int res = 0;
        rep (j , 1 , n) {
            if ( same (e[ i ] , e[ j ])) res ++;
        }
        ans = max (ans , res);
    }
    int ma = 0;
    rep (i , 1 , n) {
        rep(j , i + 1 , n) {
            if ( same (e[ i ] , e[ j ])) continue;
            ll x1 = e[ i ].c.x , y1 = e[ i ].c.y;
            ll x2 = e[ j ].c.x , y2 = e[ j ].c.y;
            ll a = y2 - y1;
            ll b = x1 - x2;
            ll c = x2 * y1 - x1 * y2;
            Point v = e[ j ].c - e[ i ].c;
            int ans1 = 2;
            int ans2 = 2;
            rep (k , 1 , n) {
                if ( k == i || k == j ) continue;
                Point v1 = e[ k ].c - e[ i ].c;
                if ( inter (a , b , c , e[ k ]) && ( v ^ v1 ) <= 0 ) ans1 ++;
                if ( inter (a , b , c , e[ k ]) && ( v ^ v1 ) >= 0 ) ans2 ++;
            }
            ma = max (ma , max (ans1 , ans2));
        }
    }
    ans = max (ma , ans);
    cout << ans << endl;
    return 0;
}