【POJ】1556 The Doors（计算几何基础+spfa）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }

const double eps=1e-6;
int dcmp(double x) { return abs(x)<eps?0:(x<0?-1:1); }
struct ipoint { double x, y; };
double icross(ipoint &a, ipoint &b, ipoint &c) {
	static double x1, x2, y1, y2;
	x1=a.x-c.x; y1=a.y-c.y;
	x2=b.x-c.x; y2=b.y-c.y;
	return x1*y2-x2*y1;
}
int ijiao(ipoint &p1, ipoint &p2, ipoint &q1, ipoint &q2) {
	return (dcmp(icross(p1, q1, q2))^dcmp(icross(p2, q1, q2)))==-2 &&
		   (dcmp(icross(q1, p1, p2))^dcmp(icross(q2, p1, p2)))==-2;
}

const int N=1000;
struct dat { int next, to; double w; }e[N<<2];
int ihead[N], cnt;
void add(int u, int v, double w) {
	e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].w=w;
}
double spfa(int s, int t, int n) {
	static double d[N];
	static int q[N], front, tail, u, v;
	static bool vis[N];
	front=tail=0;
	for1(i, 0, n) vis[i]=0, d[i]=1e99;
	d[s]=0; q[tail++]=s; vis[s]=1;
	while(front!=tail) {
		u=q[front++]; if(front==N) front=0; vis[u]=0;
		rdm(u, i) if(d[v=e[i].to]+eps>d[u]+e[i].w) {
			d[v]=d[u]+e[i].w;
			if(!vis[v]) {
				vis[v]=1;
				if(d[v]<d[q[front]]+eps) {
					--front; if(front<0) front+=N;
					q[front]=v;
				}
				else { q[tail++]=v; if(tail==N) tail=0; }
			}
		}
	}
	return d[t];
}

ipoint p[N], line[N*3][2];
int n, pn, ln;

bool check(ipoint &x, ipoint &y) {
	for1(i, 1, ln) if(ijiao(x, y, line[i][0], line[i][1])) return false;
	return true;
}
double sqr(double x) { return x*x; }
double dis(ipoint &x, ipoint &y) { return sqrt(sqr(x.x-y.x)+sqr(x.y-y.y)); }

int main() {
	while(read(n), n!=-1) {
		ln=0; pn=0;
		++pn; p[pn].x=0; p[pn].y=5;
		++pn; p[pn].x=10; p[pn].y=5;
		static double rx, ry[4];
		while(n--) {
			scanf("%lf", &rx);
			rep(k, 4) scanf("%lf", &ry[k]);
			++ln; line[ln][0]=(ipoint){rx, 0};		line[ln][1]=(ipoint){rx, ry[0]};
			++ln; line[ln][0]=(ipoint){rx, ry[1]};	line[ln][1]=(ipoint){rx, ry[2]};
			++ln; line[ln][0]=(ipoint){rx, ry[3]};	line[ln][1]=(ipoint){rx, 10};
			rep(k, 4) ++pn, p[pn].x=rx, p[pn].y=ry[k];
		}
		for1(i, 1, pn) for1(j, 1, pn) if(i!=j && check(p[i], p[j])) add(i, j, dis(p[i], p[j]));
		printf("%.2f\n", spfa(1, 2, pn));

		memset(ihead, 0, sizeof(int)*(pn+1));
		cnt=0;
	}
	return 0;
}