[haoi2014]穿越封锁线

Posted 2020-09-01 CHADLZX

 

这题需要注意的一点是射线法需要考虑边界,而且题目对边界的限制极为严格.

dcmp(v[i%n].x-x)<=0&&dcmp(v[(i+1)%n].x-x)>0
dcmp(v[i%n].x-x)>0&&dcmp(v[(i+1)%n].x-x)<=0

这是我写的版本.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iomanip>
using namespace std;
#define LL long long
#define up(i,j,n) for(int i=j;i<=n;i++)
#define pii pair<int,int>
#define db double
#define eps 1e-4
#define FILE "dealing"
int read(){
	int x=0,f=1,ch=getchar();
	while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch<=‘9‘&&ch>=‘0‘){x=(x<<1)+(x<<3)+ch-‘0‘,ch=getchar();}
	return x*f;
}
const int maxn=22000,inf=1000000000;
bool cmin(db& a,db b){return a>b?a=b,true:false;}
int n;
struct vec {
	db x,y;
	vec(db x=0,db y=0):x(x),y(y){}
}v[maxn];
int dcmp(db v){if(fabs(v)<eps)return 0;return v<0?-1:1;}
db x,y;
db s[5];
db gety(vec v,vec b,db x){
	db k=(b.y-v.y)/(b.x-v.x);
	db d=v.y-v.x*k;
	return x*k+d;
}
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	n=read();
	up(i,0,n-1)v[i].x=read(),v[i].y=read();
	x=read(),y=read();
	int wn=1;db ans=0,last=y;
	db Min=inf;int k=0;
	up(i,0,n-1)
	if(((v[i%n].x<x&&v[(i+1)%n].x>x)||(v[i%n].x>x&&v[(i+1)%n].x<x))&&gety(v[i%n],v[(i+1)%n],x)>y&&cmin(Min,gety(v[i],v[(i+1)%n],x)-y)){k=i;}
	up(i,k,k+n-1){
		if(dcmp(v[i%n].x-x)<=0&&dcmp(v[(i+1)%n].x-x)>0&&dcmp(gety(v[i%n],v[(i+1)%n],x)-y)>0){
			s[wn]+=gety(v[i%n],v[(i+1)%n],x)-last;
			last=gety(v[i%n],v[(i+1)%n],x);
			wn++;
		}
		else if(dcmp(v[i%n].x-x)>0&&dcmp(v[(i+1)%n].x-x)<=0&&dcmp(gety(v[i%n],v[(i+1)%n],x)-y)>0){
			s[wn]+=gety(v[i%n],v[(i+1)%n],x)-last;
			last=gety(v[i%n],v[(i+1)%n],x);
			wn--;
		}
	}
	if(wn==1)printf("%d\n",(int)fabs(s[2]+s[0]));
	else printf("%d\n",fabs(s[1]));
	return 0;
}