OJ-Problems-Source/HDOJ/4697_autoAC.cpp

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-9;
inline int dcmp(double x) {
    if (fabs(x) < eps)
        return 0;
    return x > eps ? 1 : -1;
}
struct point {
    double x, y;
    int id;
    inline void in(int i) {
        scanf("%lf%lf", &x, &y);
        id = i;
    }
    point(double x = 0, double y = 0) :
            x(x), y(y) {
    }
    inline point operator -(const point &t) const {
        return point(x - t.x, y - t.y);
    }
    inline point operator +(const point &t) const {
        return point(x + t.x, y + t.y);
    }
    point operator /(const double &t) const {
        return point(x / t, y / t);
    }
    point operator *(const double &t) const {
        return point(x * t, y * t);
    }
    inline double operator *(const point &t) const {            
        return x*t.y-y*t.x;
    }
    bool operator <(const point &t) const {
        if(y == t.y) return x < t.x;
        return y < t.y;
    }
}p[55], v[55], tp[55], st[55];
double t[55*55*55], T, cur, ans;
int sz, n, m;
double a, b, c;
inline double cross(const point &o, const point &a, const point &b) {
    return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
int graham(point *p, int n, point *st) {        
    sort(p, p + n);
    int top = 0;
    int i;
    for (i = 0; i < n; i++) {
        while (top >= 2 && cross(st[top - 2], st[top - 1], p[i]) < eps)
            top--;
        st[top++] = p[i];
    }
    int t = top + 1;
    for (i = n - 2; i >= 0; i--) {
        while (top >= t && cross(st[top - 2], st[top - 1], p[i]) < eps)
            top--;
        st[top++] = p[i];
    }
    return top;
}
inline void gao(int i, int j, int k) {        
    point a1 = p[i]-p[j], a2 = p[i]-p[k];
    point b1 = v[i]-v[j], b2 = v[i]-v[k];
    a += b1*b2; b += a1*b2+b1*a2; c += a1*a2;
}
inline void solve(double a, double b, double c) {    
    double x;
    if(a == 0) {
        if(b!= 0) {
            x = -c/b;
            if(x >= 0 && x <= T) t[sz++] = x;
        }
        return;
    }
    double dlt = b*b-4*a*c;
    if(dlt < 0) return;
    if(dlt == 0) {
        x = -b*0.5/a;
        if(x >= 0 && x <= T) t[sz++] = x;
        return;
    }
    dlt = sqrt(dlt);
    x = 0.5*(-b+dlt)/a;
    if(x >= 0 && x <= T) t[sz++] = x;
    x = 0.5*(-b-dlt)/a;
    if(x >= 0 && x <= T) t[sz++] = x;
}
inline double F(double x) {        
    return a*x*x*x/3.0+b*x*x/2.0+c*x;
}
int main() {
    int i, j, k;
    while( ~scanf("%d%lf", &n, &T)) {
        for(i = 0; i < n; i++) p[i].in(i), v[i].in(i);
        if(n <= 2) {
            printf("%.10f\n", 0.0);
            continue;
        }
        t[0] = 0; t[1] = T;    
        sz = 2;
        for(i = 0; i < n; i++)
            for(j = i+1; j < n; j++)
                for(k = j+1; k < n; k++) {
                    a = b = c = 0;
                    gao(i, j, k);
                    solve(a, b, c);
                }
        sort(t, t+sz);
        ans = 0;
        for(i = 0; i < sz-1; i++) {
            cur = 0.5*(t[i]+t[i+1]);
            a = b = c = 0;
            for(j = 0; j < n; j++) {
                tp[j] = p[j]+v[j]*cur;
                tp[j].id = p[j].id;
            }
            m = graham(tp, n, st);
            for(j = 2; j < m; j++)
                gao(st[0].id, st[j-1].id, st[j].id);    
            ans += F(t[i+1])- F(t[i]);
        }
        printf("%.10f\n", fabs(ans*0.5/T));
    }
    return 0;
}