链接：https://ac.nowcoder.com/acm/contest/12548/I
来源：牛客网
题目描述
Mr. Main took a train from city ss to city tt and passed a plain full of windmills. The train ran in a straight line. A windmill is a machine used for wind power generation. Its fan blades rotate when the wind blows. From his perspective, colorful windmills lined up on the horizon from left to right.
As the train was running, the order of windmills from his perspective was constantly changing: a windmill was originally on the left/right of another, and then changed to its right/left;
Given the coordinates of the windmills, please find the coordinate of him when he just observed the hh-th windmill exchanged order with other windmills for the kk-th times. It is guaranteed that any three of the points given, the cities and the windmills, were not collinear, and that all of the windmills were on the same side of the line that the train ran along.
As shown in the picture, in Mr. Mian's perspective, B was initially to the left of A, and later to the right of A.
输入描述:
The first line of input contains two integers nn and mm, where n(1≤n≤1000)n(1≤n≤1000) is number of windmills, and m(1≤m≤104)m(1≤m≤104) is number of queries.
The second line contains four integers xsxs, ysys, xtxt and ytyt (−106≤xs,ys,xt,yt≤106−106≤xs,ys,xt,yt≤106), which are the coordinates of the starting city s and destination city t.
The next nn lines describe the windmills, the ii-th of which contains two integers xixi,
yiyi (−106≤xi,yi≤106−106≤xi,yi≤106), which are the coordinates of the ii-th windmill.
The next mm lines describe the queries, the ii-th of which contains two integers, hihi and kiki (1≤hi≤n,1≤ki≤1061≤hi≤n,1≤ki≤106), denoting a query for the coordinates when observing the kiki-th pass of the hihi-th windmill.
输出描述:
Output m lines, each containing two real numbers xi,yixi,yi, representing the coordinates when the hihi-th windmill is observed to exchange order with other windmills for k times; if it does not exist, output -1. Your answer is considered correct if its absolute or relative error with the standard answer is less than 10−510−5.
示例1
输入
复制
4 2
0 0 5 0
1 3
2 4
4 1
4 5
1 2
3 2
输出
复制
-1
4.6666666667 0.0000000000
读懂题后看到数据范围发现其实就是暴力。对于某个点h，它和另一个点g在观察者的视角发生位置左右置换的临界位置就是hg这条直线与st这条线段的交点（如果有的话）。因此对于每个点h，求出它和另外所有点构成的直线与st线段的交点扔进vector然后进行排序（排序原则是点到s的距离近的在前），查询的时候如果有解（对于h这个点的vector的size大与等于当前查询的k）则直接输出v[h - 1, k - 1]，否则输出-1。
板子可以直接用kuangbin的计算几何板子，精度设置为1e-8貌似没啥问题。
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
int n, m;
double xs, ys, xt, yt, h, k;
double x[1005], y[1005];
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//`Compares a double to zero`
int sgn(double x){
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
//square of a double
inline double sqr(double x){return x*x; }
struct Point{
double x,y;
Point(){}
Point(double _x,double _y){
x = _x;
y = _y;
}
void input(){
scanf("%lf%lf",&x,&y);
}
void output(){
printf("%.2f %.2f
",x,y);
}
bool operator == (Point b)const{
return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
}
bool operator < (Point b)const{
return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;
}
Point operator -(const Point &b)const{
return Point(x-b.x,y-b.y);
}
//叉积
double operator ^(const Point &b)const{
return x*b.y - y*b.x;
}
//点积
double operator *(const Point &b)const{
return x*b.x + y*b.y;
}
//返回长度
double len(){
return hypot(x,y);//库函数
}
//返回长度的平方
double len2(){
return x*x + y*y;
}
//返回两点的距离
double distance(Point p){
return hypot(x-p.x,y-p.y);
}
Point operator +(const Point &b)const{
return Point(x+b.x,y+b.y);
}
Point operator *(const double &k)const{
return Point(x*k,y*k);
}
Point operator /(const double &k)const{
return Point(x/k,y/k);
}
//`计算pa 和 pb 的夹角`
//`就是求这个点看a,b 所成的夹角`
//`测试 LightOJ1203`
double rad(Point a,Point b){
Point p = *this;
return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
}
//`化为长度为r的向量`
Point trunc(double r){
double l = len();
if(!sgn(l))return *this;
r /= l;
return Point(x*r,y*r);
}
//`逆时针旋转90度`
Point rotleft(){
return Point(-y,x);
}
//`顺时针旋转90度`
Point rotright(){
return Point(y,-x);
}
//`绕着p点逆时针旋转angle`
Point rotate(Point p,double angle){
Point v = (*this) - p;
double c = cos(angle), s = sin(angle);
return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);
}
};
struct Line{
Point s,e;
Line(){}
Line(Point _s,Point _e){
s = _s;
e = _e;
}
bool operator ==(Line v){
return (s == v.s)&&(e == v.e);
}
//`根据一个点和倾斜角angle确定直线,0<=angle<pi`
Line(Point p,double angle){
s = p;
if(sgn(angle-pi/2) == 0){
e = (s + Point(0,1));
}
else{
e = (s + Point(1,tan(angle)));
}
}
//ax+by+c=0
Line(double a,double b,double c){
if(sgn(a) == 0){
s = Point(0,-c/b);
e = Point(1,-c/b);
}
else if(sgn(b) == 0){
s = Point(-c/a,0);
e = Point(-c/a,1);
}
else{
s = Point(0,-c/b);
e = Point(1,(-c-a)/b);
}
}
void input(){
s.input();
e.input();
}
void adjust(){
if(e < s)swap(s,e);
}
//求线段长度
double length(){
return s.distance(e);
}
//`返回直线倾斜角 0<=angle<pi`
double angle(){
double k = atan2(e.y-s.y,e.x-s.x);
if(sgn(k) < 0)k += pi;
if(sgn(k-pi) == 0)k -= pi;
return k;
}
//`点和直线关系`
//`1 在左侧`
//`2 在右侧`
//`3 在直线上`
int relation(Point p){
int c = sgn((p-s)^(e-s));
if(c < 0)return 1;
else if(c > 0)return 2;
else return 3;
}
// 点在线段上的判断
bool pointonseg(Point p){
return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
}
//`两向量平行(对应直线平行或重合)`
bool parallel(Line v){
return sgn((e-s)^(v.e-v.s)) == 0;
}
//`两线段相交判断`
//`2 规范相交`
//`1 非规范相交`
//`0 不相交`
int segcrossseg(Line v){
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
int d3 = sgn((v.e-v.s)^(s-v.s));
int d4 = sgn((v.e-v.s)^(e-v.s));
if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
}
//`直线和线段相交判断`
//`-*this line -v seg`
//`2 规范相交`
//`1 非规范相交`
//`0 不相交`
int linecrossseg(Line v){
int d1 = sgn((e-s)^(v.s-s));
int d2 = sgn((e-s)^(v.e-s));
if((d1^d2)==-2) return 2;
return (d1==0||d2==0);
}
//`两直线关系`
//`0 平行`
//`1 重合`
//`2 相交`
int linecrossline(Line v){
if((*this).parallel(v))
return v.relation(s)==3;
return 2;
}
//`求两直线的交点`
//`要保证两直线不平行或重合`
Point crosspoint(Line v){
double a1 = (v.e-v.s)^(s-v.s);
double a2 = (v.e-v.s)^(e-v.s);
return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
}
//点到直线的距离
double dispointtoline(Point p){
return fabs((p-s)^(e-s))/length();
}
//点到线段的距离
double dispointtoseg(Point p){
if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
return min(p.distance(s),p.distance(e));
return dispointtoline(p);
}
//`返回线段到线段的距离`
//`前提是两线段不相交，相交距离就是0了`
double dissegtoseg(Line v){
return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
}
//`返回点p在直线上的投影`
Point lineprog(Point p){
return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
}
//`返回点p关于直线的对称点`
Point symmetrypoint(Point p){
Point q = lineprog(p);
return Point(2*q.x-p.x,2*q.y-p.y);
}
};
Point ss;
bool cmp(Point a, Point b) {
return abs(a.x - ss.x) < abs(b.x - ss.x);
}
int main() {
cin >> n >> m;
vector<vector<Point> > v;
cin >> xs >> ys >> xt >> yt;
for(int i = 1; i <= n; i++) {
cin >> x[i] >> y[i];
}
Point start(xs, ys);
ss = start;
Point end(xt, yt);
Line road(start, end);
for(int i = 1; i <= n; i++) {
vector<Point> tmp;
//v.push_back(tmp);
for(int j = 1; j <= n; j++) {
if(i == j) continue;
Line now(Point{x[i], y[i]}, Point{x[j], y[j]});
if(now.linecrossseg(road)) {
tmp.push_back(now.crosspoint(road));//如果路径线段和连线直线有交点 那么直接求两直线交点即可。
}
}
sort(tmp.begin(), tmp.end(), cmp);
v.push_back(tmp);
}
for(int i = 1; i <= m; i++) {
cin >> h >> k;
if(k > v[h - 1].size()) cout << -1 << endl;
else cout << v[h - 1][k - 1].x << ' ' << v[h - 1][k - 1].y << endl;
}
return 0;
}