POJ1474 Video Surveillance(半平面交)
求多边形核的存在性，过了这题但是过不了另一题的，不知道是模板的问题还是什么，但是这个模板还是可以过绝大部分的题的。。。
#pragma warning(disable:4996)
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
#define maxn 2500
#define eps 1e-7
int n;
int dcmp(double x){
return x<-eps ? -1 : x>eps;
}
struct Point
{
double 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);
}
Point operator - (const Point &b) const{
return Point(x - b.x, y - b.y);
}
Point operator *(double d) const{
return Point(x*d, y*d);
}
Point operator /(double d) const{
return Point(x / d, y / d);
}
double det(const Point &b) const{
return x*b.y - y*b.x;
}
double dot(const Point &b) const{
return x*b.x + y*b.y;
}
Point rot90(){
return Point(-y, x);
}
Point norm(){
double len = sqrt(this->dot(*this));
return Point(x, y) / len;
}
void read(){
scanf("%lf%lf", &x, &y);
}
};
#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) (dcmp(cross(p1,p2,p3)))
Point isSS(Point p1, Point p2, Point q1, Point q2){
double a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
return (p1*a2 + p2*a1) / (a1 + a2);
}
struct Border
{
Point p1, p2;
double alpha;
void setAlpha(){
alpha = atan2(p2.y - p1.y, p2.x - p1.x);
}
};
bool operator < (const Border &a, const Border &b) {
int c = dcmp(a.alpha - b.alpha);
if (c != 0) {
return c > 0;
}
else {
return crossOp(b.p1, b.p2, a.p1) > 0;
}
}
bool operator == (const Border &a, const Border &b){
return dcmp(a.alpha - b.alpha) == 0;
}
Point isBorder(const Border &a, const Border &b){
return isSS(a.p1, a.p2, b.p1, b.p2);
}
Border border[maxn];
Border que[maxn];
int qh, qt;
// check函数判断的是新加的半平面和由a,b两个半平面产生的交点的方向，若在半平面的左侧返回True
bool check(const Border &a, const Border &b, const Border &me){
Point is = isBorder(a, b);
return crossOp(me.p1, me.p2, is) >= 0;
}
bool isParellel(const Border &a, const Border &b){
return dcmp((a.p2 - a.p1).det(b.p2 - b.p1)) == 0;
}
bool convexIntersection()
{
qh = qt = 0;
sort(border, border + n);
n = unique(border, border + n) - border;
for (int i = 0; i < n; i++){
Border cur = border[i];
while (qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], cur)) --qt;
while (qh + 1 < qt&&!check(que[qh], que[qh + 1], cur)) ++qh;
que[qt++] = cur;
}
while (qh + 1 < qt&&!check(que[qt - 2], que[qt - 1], que[qh])) --qt;
while (qh + 1 < qt&&!check(que[qh], que[qh + 1], que[qt - 1])) ++qh;
return qt - qh > 2;
}
Point ps[maxn];
int main()
{
int ca = 0;
while (cin >> n&&n)
{
for (int i = 0; i < n; i++){
ps[i].read();
}
ps[n] = ps[0];
for (int i = 0; i < n; i++){
border[i].p1 = ps[i + 1];
border[i].p2 = ps[i];
border[i].setAlpha();
}
printf("Floor #%d\n", ++ca);
if (convexIntersection()) {
puts("Surveillance is possible.");
}
else puts("Surveillance is impossible.");
puts("");
}
return 0;
}