POJ 1556 The Doors(计算几何+最短路)

Description:

You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length.

Input:

The input data for the illustrated chamber would appear as follows.

2
4 2 7 8 9
7 3 4.5 6 7

The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input file will contain at least one such set of data. The end of the data comes when the number of walls is -1.

Output:

The output should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.

Sample Input:

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

Sample Output:

10.00
10.06

题目链接

题目要求起点(0,5)到终点(10,5)的最短距离，对每两个可行路径建边，权值为其距离，建图之后求起点到终点的最短路即可。

AC代码:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;

const double INF = 1e20;
const int maxn = 1e4 + 5;

struct Point {
	double X, Y;

	Point operator - (const Point &B) const {
		return Point {X - B.X, Y - B.Y};
	}

	double operator * (const Point &B) const {
		return X * B.X + Y * B.Y;
	}

	double operator ^ (const Point &B) const {
		return X * B.Y - Y * B.X;
	}
};

struct Segment {
	Point S, T;

	double operator ^ (const Segment &B) const {
		return (T - S) ^ (B.T - B.S);
	}
};

struct Edge {
	int V;
	double Weight;
	int Next;
};

Edge edges[maxn << 2];
int Head[maxn];
int Tot;
int N;
double X, Y1, Y2, Y3, Y4;
Point points[maxn << 2];
double Dis[maxn];

void Init() {
	Tot = 0;
	memset(Head, -1, sizeof(Head));
}

void AddEdge(int U, int V, double Weight) {
	edges[Tot] = Edge {V, Weight, Head[U]};
	Head[U] = Tot++;
	edges[Tot] = Edge {U, Weight, Head[V]};
	Head[V] = Tot++;
}

double Distance(Point A, Point B) {
	return sqrt((B - A) * (B - A));
}

bool IsIntersect(Segment A, Segment B) {
	return 
		(A ^ Segment {A.S, B.S}) * (A ^ Segment {A.S, B.T}) <= 0 &&
		(B ^ Segment {B.S, A.S}) * (B ^ Segment {B.S, A.T}) <= 0;
}

struct Cmp {
	bool operator () (const int &A, const int &B) {
		return Dis[A] > Dis[B];
	}
};

void Dijkstra(int Start) {
	priority_queue<int, vector<int>, Cmp> Que;
	for (int i = 0; i <= 4 * N + 1; ++i) {
		Dis[i] = INF;
	}
	Dis[Start] = 0;
	Que.push(Start);
	while (!Que.empty()) {
		int Cur = Que.top(); Que.pop();
		for (int i = Head[Cur]; ~i; i = edges[i].Next) {
			if (Dis[edges[i].V] > Dis[Cur] + edges[i].Weight) {
				Dis[edges[i].V] = Dis[Cur] + edges[i].Weight;
				Que.push(edges[i].V);
			}
		}
	}
}

int main(int argc, char *argv[]) {
	while (~scanf("%d", &N) && ~N) {
		Init();

		points[0] = Point {0, 5};
		for (int i = 1; i <= N; ++i) {
			scanf("%lf%lf%lf%lf%lf", &X, &Y1, &Y2, &Y3, &Y4);
			points[i * 4 - 3] = Point {X, Y1};
			points[i * 4 - 2] = Point {X, Y2};
			points[i * 4 - 1] = Point {X, Y3};
			points[i * 4] = Point {X, Y4};
		}
		points[4 * N + 1] = Point {10, 5};

		for (int i = 1; i <= 4 * N; ++i) {
			bool Flag = true;
			Segment Judge = Segment {points[0], points[i]};
			for (int j = 4; j < i; j += 4) {
				double CurX = points[j].X;
				if (IsIntersect(Judge, Segment {Point {CurX, 0}, points[j - 3]}) ||
					IsIntersect(Judge, Segment {points[j - 2], points[j - 1]}) ||
					IsIntersect(Judge, Segment {points[j], Point {CurX, 10}})) {
					Flag = false;
					break;
				}
			}
			if (Flag) {
				AddEdge(0, i, Distance(points[0], points[i]));
			}

			Flag = true;
			Judge = Segment {points[i], points[4 * N + 1]};
			for (int j = 4 * N; j - 3 > i; j -= 4) {
				double CurX = points[j].X;
				if (IsIntersect(Judge, Segment {Point {CurX, 0}, points[j - 3]}) ||
					IsIntersect(Judge, Segment {points[j - 2], points[j - 1]}) ||
					IsIntersect(Judge, Segment {points[j], Point {CurX, 10}})) {
					Flag = false;
					break;
				}
			}
			if (Flag) {
				AddEdge(i, 4 * N + 1, Distance(points[i], points[4 * N + 1]));
			}
		}

		for (int i = 1; i <= 4 * N; ++i) {
			for (int l = 4 * N; l - 3 > i; l -= 4) {
				for (int j = l; j >= l - 3; --j) {
					bool Flag = true;
					Segment Judge = Segment {points[i], points[j]};
					for (int k = l - 4; k - 3 > i; k -= 4) {
						double CurX = points[k].X;
						if (IsIntersect(Judge, Segment {Point {CurX, 0}, points[k - 3]}) ||
							IsIntersect(Judge, Segment {points[k - 2], points[k - 1]}) ||
							IsIntersect(Judge, Segment {points[k], Point {CurX, 10}})) {
							Flag = false;
							break;
						}
					}
					if (Flag) {
						AddEdge(i, j, Distance(points[i], points[j]));
					}
				}
			}
		}

		bool Flag = true;
		Segment Judge = Segment {points[0], points[4 * N + 1]};
		for (int j = 4; j <= 4 * N; j += 4) {
			double CurX = points[j].X;
			if (IsIntersect(Judge, Segment {Point {CurX, 0}, points[j - 3]}) ||
				IsIntersect(Judge, Segment {points[j - 2], points[j - 1]}) ||
				IsIntersect(Judge, Segment {points[j], Point {CurX, 10}})) {
				Flag = false;
				break;
			}
		}
		if (Flag) {
			AddEdge(0, 4 * N + 1, Distance(points[0], points[4 * N + 1]));
		}

		Dijkstra(0);
		printf("%.2f\n", Dis[4 * N + 1]);
	}
	return 0;
}