BZOJ 3800 Saber VS Lancer/POJ 1755 Triathlon semi-plane intersection, bzojpoj

BZOJ 3800 Saber VS Lancer/POJ 1755 Triathlon semi-plane intersection, bzojpoj

N people have a fixed speed for these three sports, but the length of each sport is unknown. Now, I am wondering if the referee can adjust the length of the three sports to win a single player.

Thought: Now, when we want to win a person, the three speeds for this person are v1, v2, and v3, respectively, and we want everyone to lose to him, set the speed of a person to v1, v2, and v3 '. Set the length of the three matches to l1, l2, and l3. It is not difficult to obtain the following equation: l1/v1 + l2/v2 + l3/v3 <l1/v1 '+ l2/v2' + l3/v3 '.

But this equation has three unknowns. What is this? It cannot be processed at all. In fact, this question only requires us to determine whether there is a solution, so we do not pay attention to the specific value. If a solution of X, Y, and Z allows a person to win everyone, then X * K, Y * K, and Z * K can also win everyone. Therefore, determining an l value does not affect the correctness of the answer. So suppose l3 is 1, then l3/v3 and l3/v3 are two values.

The original equation becomes: (for ease of understanding, set x = l1, y = l2) x * (v1 '-v1)/(v1 * v1 ') + y * (v2 '-v2)/(v2 * v2') + k-k' <0.

In this formula, we only do not know xy, and the rest are constants. This is a half plane of x and y. Finding all the Half Planes and determining whether the kernel exists is the answer.

The data range is very small. We can use the semi-plane intersection of O (n ^ 2.

However, I did not have this question .. The accuracy of this question card is simply unbearable. After obtaining data from the Internet, we use cena for testing. There are a total of 25 points, that is, there are two points that cannot survive or survive .. In my heart, AC ..

CODE (not AC, submit with caution ):

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 10010#define EPS 1e-16#define INF (1ll << 32)using namespace std;#define DCMP(a) (fabs(a) < EPS)  struct Complex{    double v1,v2,v3;          void Read() {        scanf("%lf%lf%lf",&v1,&v2,&v3);    }}src[MAX];  struct Point{    double x,y;          Point(double _ = .0,double __ = .0):x(_),y(__) {}    Point operator +(const Point &a)const {        return Point(x + a.x,y + a.y);    }    Point operator -(const Point &a)const {        return Point(x - a.x,y - a.y);    }    Point operator *(double a)const {        return Point(x * a,y * a);    }}p[MAX];  struct Line{    Point p,v;    double alpha;          bool operator <(const Line &a)const {        return alpha < a.alpha;    }    Line(Point _,Point __):p(_),v(__) {        alpha = atan2(v.y,v.x);    }    Line() {}}line[MAX],q[MAX];  int cnt,lines;  inline double Cross(const Point &a,const Point &b){    return a.x * b.y - a.y * b.x;}  inline bool OnLeft(const Line &l,const Point &p){    return Cross(l.v,p - l.p) > 0;}  inline void MakeLine(const Complex &win,const Complex &lose){    double A = (lose.v1 - win.v1) / (win.v1 * lose.v1);    double B = (lose.v2 - win.v2) / (win.v2 * lose.v2);    double C = (lose.v3 - win.v3) / (win.v3 * lose.v3);    if(DCMP(A) && DCMP(B) && C < EPS)        throw false;    if(DCMP(B)) B = EPS;    line[++lines] = Line(Point(.0,-C / B),Point(-B,A));}  inline void Initialize(){    line[++lines] = Line(Point(0,0),Point(1,0));    line[++lines] = Line(Point(INF,0),Point(0,1));    line[++lines] = Line(Point(INF,INF),Point(-1,0));    line[++lines] = Line(Point(0,INF),Point(0,-1)); }  inline Point GetIntersection(const Line &a,const Line &b){    Point u = a.p - b.p;    double temp = Cross(b.v,u) / Cross(a.v,b.v);    return a.p + a.v * temp;}  inline bool HalfplaneIntersection(){    int front = 1,tail = 1;    q[1] = line[1];    for(int i = 2;i <= lines; ++i) {        while(front < tail && !OnLeft(line[i],p[tail - 1]))  --tail;        while(front < tail && !OnLeft(line[i],p[front])) ++front;        if(DCMP(Cross(line[i].v,q[tail].v)))            q[tail] = OnLeft(q[tail],line[i].p) ? line[i]:q[tail];        else    q[++tail] = line[i];        if(front < tail) p[tail - 1] = GetIntersection(q[tail],q[tail - 1]);    }    while(front < tail && !OnLeft(q[front],p[tail - 1])) --tail;    if(tail - front <= 1)    return false;    double re = Cross(tail,front);    for(int i = front; i < tail; ++i)        re += Cross(p[i],p[i + 1]);    return !DCMP(re);}  int main(){    cin >> cnt;    for(int i = 1; i <= cnt; ++i)        src[i].Read();    for(int i = 1; i <= cnt; ++i) {        lines = 0;        try{            for(int j = 1; j <= cnt; ++j) {                if(i == j)  continue;                MakeLine(src[i],src[j]);            }        }        catch(bool flag) {            puts("No");            continue;        }        Initialize();        sort(line + 1,line + lines + 1);        if(HalfplaneIntersection()) puts("Yes");        else    puts("No");    }    return 0;}