POJ 1474 Video Surveillance （多边形内核判断）

/*
* Problem: POJ No.1474
* Running time: 125MS
* Complier: G++
* Author: herongwei
* Create Time: 12:27 2015/10/2 星期五
*/

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi=acos(-1.0);
const double e=exp(1.0);
const double eps=1e-8;
const int maxn=105;

struct Point{
double x,y;
}point[maxn];

Point temp[maxn];
Point p[maxn];
int pre_point ,last_point;
double a,b,c;

void getline(Point x,Point y)//获取直线ax+by+c==0
{
a=y.y-x.y;
b=x.x-y.x;
c=y.x*x.y-x.x*y.y;
}

Point intersect(Point x,Point y)//获取直线ax+by+c==0  和点x和y所连直线的交点
{
double u=fabs(a*x.x+b*x.y+c);
double v=fabs(a*y.x+b*y.y+c);
Point ans;
ans.x=(x.x*v+y.x*u)/(u+v);
ans.y=(x.y*v+y.y*u)/(u+v);
return ans;
}

void cut()//用直线ax+by+c==0切割多边形
{
int cut_num=0;
for(int i=1; i<=last_point; ++i)
{
if(a*p[i].x+b*p[i].y+c>=0){
temp[++cut_num]=p[i];
}
else
{
if(a*p[i-1].x+b*p[i-1].y+c>0)
{
temp[++cut_num]=intersect(p[i-1],p[i]);
}
if(a*p[i+1].x+b*p[i+1].y+c>0)
{
temp[++cut_num]=intersect(p[i+1],p[i]);
}
}
}
for(int i=1; i<=cut_num; ++i)
{
p[i]=temp[i];
}
p[cut_num+1]=temp[1];
p[0]=temp[cut_num];
last_point=cut_num;
}

void solve()
{
for(int i=1; i<=pre_point; ++i){
p[i]=point[i];
}
point[pre_point+1]=point[1];
p[pre_point+1]=p[1];
p[0]=p[pre_point];
last_point=pre_point;
for(int i=1; i<=pre_point; ++i)
{
getline(point[i],point[i+1]);//根据point[i]和point[i+1]确定直线ax+by+c==0
cut();//用直线ax+by+c==0切割多边形
}
}
int main()
{
int tot=1;
while(cin>>pre_point&&pre_point){
for(int i=1; i<=pre_point; ++i){
cin>>point[i].x>>point[i].y;
}
solve();
printf("Floor #%d\n",tot++);
if(last_point==0) puts("Surveillance is impossible."),puts("");
else puts("Surveillance is possible."),puts("");
}
return 0;
}
/*
4
0 0
0 1
1 1
1 0
8
0 0
0 2
1 2
1 1
2 1
2 2
3 2
3 0
0
Floor #1
Surveillance is possible.

Floor #2
Surveillance is impossible.
*/