POJ 1556The Doors 判断线段相交+SPFA

Posted 2020-07-02 abclzr

黑书上的一道例题：如果走最短路则会碰到点，除非中间没有障碍。

这样把能一步走到的点两两连边，然后跑SPFA即可。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 100003
using namespace std;
struct Point {
	double x, y;
	Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
};
inline int dcmp(double a) {
	return (fabs(a) < 1e-6) ? 0 : (a < 0 ? -1 : 1);
}
Point operator - (Point a, Point b) {
	return Point(a.x - b.x, a.y - b.y);
}
bool operator == (Point a, Point b) {
	return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
inline double Cross(Point a, Point b) {
	return a.x * b.y - a.y * b.x;
}
inline double Dot(Point a, Point b) {
	return a.x * b.x + a.y * b.y;
}
inline bool jiao(Point d1, Point d2, Point d3, Point d4) {
	return (dcmp(Cross(d4 - d3, d1 - d3)) ^ dcmp(Cross(d4 - d3, d2 - d3))) == -2 &&
			(dcmp(Cross(d2 - d1, d3 - d1)) ^ dcmp(Cross(d2 - d1, d4 - d1))) == -2;
}
inline int bjiao(Point d1, Point d2, Point d3) {
	if (d1 == d2 || d1 == d3)
		return 1;
	if (dcmp(Cross(d2 - d1, d3 - d1)) == 0 && dcmp(Dot(d2 - d1, d3 - d1)) == -1)
		return 1;
	return 0;
}
inline double dis(Point d1, Point d2) {
	return sqrt((d2.y - d1.y) * (d2.y - d1.y) + (d2.x - d1.x) * (d2.x - d1.x));
}

struct nodeline {
	Point a, b;
	nodeline(Point _a = (0, 0) , Point _b = (0, 0)) : a(_a), b(_b) {}
} line[N];

struct node {
	int nxt, to;
	double w;
} E[N << 1];

Point a[N];
int n, cnt, lcnt, point[N], dd, q[N];
double nowx, nowy, nowy2, dist[N];
bool vis[N];

inline void ins(int x, int y, double z) {++dd; E[dd].nxt = point[x]; E[dd].to = y; E[dd].w = z; point[x] = dd;}
inline double spfa(int S, int T) {
	memset(dist, 127, sizeof(dist));
	memset(vis, 0, sizeof(vis));
	int head = 0, tail = 1;
	q[1] = S;
	vis[S] = 1;
	dist[S] = 0;
	while (head != tail) {
		++head; if (head >= N) head %= N;
		int u = q[head];
		vis[u] = 0;
		for(int tmp = point[u]; tmp; tmp = E[tmp].nxt) {
			int v = E[tmp].to;
			if (dist[u] + E[tmp].w < dist[v]) {
				dist[v] = dist[u] + E[tmp].w;
				if (!vis[v]) {
					++tail; if (tail >= N) tail %= N;
					q[tail] = v;
					vis[v] = 1;
				}
			}
		}
	}
	return dist[T];
}

inline bool check(Point d1, Point d2) {
	for(int i = 1; i <= lcnt; ++i)
		if (jiao(d1, d2, line[i].a, line[i].b))
			return 0;
	return 1;
}

int main() {
	scanf("%d", &n);
	while (n != -1) {
		cnt = 2;
		lcnt = 0;
		a[1].x = 0;
		a[1].y = 5;
		a[2].x = 10;
		a[2].y = 5;
		for(int i = 1; i <= n; ++i) {
			scanf("%lf", &nowx);
			scanf("%lf", &nowy);
			a[++cnt] = Point(nowx, nowy);
			line[++lcnt] = nodeline(Point(nowx, 0), a[cnt]);
			scanf("%lf", &nowy);
			a[++cnt] = Point(nowx, nowy);
			scanf("%lf", &nowy2);
			a[++cnt] = Point(nowx, nowy2);
			line[++lcnt] = nodeline(a[cnt - 1], a[cnt]);
			scanf("%lf", &nowy);
			a[++cnt] = Point(nowx, nowy);
			line[++lcnt] = nodeline(a[cnt], Point(nowx, 10));
		}
		
		memset(point, 0 , sizeof(point));
		dd = 0;
		for(int i = 1; i <= cnt; ++i)
			for(int j = i + 1; j <= cnt; ++j)
				if (check(a[i], a[j]))
					ins(i, j, dis(a[i], a[j]));
		for(int i = 1; i <= cnt; ++i)
			if (i != 2 && check(a[i], a[2]))
				ins(i, 2, dis(a[i], a[2]));
		printf("%.2lf\n", spfa(1,2));
		scanf("%d", &n);
	}
	return 0;
}

清明节机房也放假啊滚来滚去……~(～o￣▽￣)～o 。。。滚来滚去……o～(＿△＿o～) ~。。。