【计算几何】 POJ 1556 The Doors

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和图论中的最短路密切相关的一道题目，把所有的线段端点看做一个无向图中的顶点，两个顶点如果能够通过线段连接（需要判断线段相交，经典的计算几何基础），则路径长度为线段长度，否则置为无穷大。在构造的图上运行最短路算法就可以了（Dijkstra算法就行，不过floyd算法更容易实现，尽管复杂度略高。）

#include <vector>#include <list>#include <limits.h>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <string.h>#include <stdlib.h>#include <cassert>using namespace std;struct Point {    double x, y;    Point(double _x, double _y): x(_x), y(_y) {}    Point() {}};struct Line {    Point p1, p2;    Line(Point _p1, Point _p2): p1(_p1), p2(_p2) {}    Line() {}};double cross(Point a, Point b1, Point b2) {    return (b1.x - a.x) * (b2.y - a.y) - (b1.y - a.y) * (b2.x - a.x);}bool intersect(Line a, Line b) {    Point pa1 = a.p1, pa2 = a.p2;    Point pb1 = b.p1, pb2 = b.p2;    return max(pa1.x, pa2.x) > min(pb1.x, pb2.x)        && max(pb1.x, pb2.x) > min(pa1.x, pa2.x)        && max(pa1.y, pa2.y) > min(pb1.y, pb2.y)        && max(pb1.y, pb2.y) > min(pa1.y, pa2.y)        && cross(pa1, pa2, pb1) * cross(pa1, pa2, pb2) < 0        && cross(pb1, pb2, pa1) * cross(pb1, pb2, pa2) < 0; }double euclidean(Point p1, Point p2) {    return sqrt((p1.x - p2.x) * (p1.x - p2.x)                 + (p1.y - p2.y) * (p1.y - p2.y));}vector<Line> v;vector<Point> ps;const int MAX_N = 80;double dis[MAX_N][MAX_N];int main() {    int N, M;    while (cin >> N && N >= 0) {        v.clear(); ps.clear();        ps.push_back(Point(0, 5));        ps.push_back(Point(10, 5));        double x, y;        Point p[4];        for (int i = 0; i < N; ++i) {            cin >> x;            for (int j = 0; j < 4; ++j) { cin >> y; p[j] = Point(x, y); ps.push_back(Point(x, y)); }            v.push_back(Line(Point(x, 0), p[0]));            v.push_back(Line(p[1], p[2]));            v.push_back(Line(p[3], Point(x, 10)));        }        int T = ps.size();        for (int i = 0; i < T; ++i)        for (int j = 0; j < T; ++j) {            if (i == j) dis[i][j] = 0;            else dis[i][j] = INT_MAX;        }        for (int i = 0; i < T; ++i)        for (int j = 0; j < i; ++j) {            Line l(ps[i], ps[j]);            bool ok = true;            for (int k = 0; k < v.size(); ++k) {                if (intersect(v[k], l)) { ok = false; break; }            }            if (ok) {                dis[i][j] = min(dis[i][j], euclidean(ps[i], ps[j]));                dis[j][i] = min(dis[j][i], euclidean(ps[i], ps[j]));            }        }        // floyd is easier to implement        for (int k = 0; k < T; ++k)        for (int i = 0; i < T; ++i)        for (int j = 0; j < T; ++j) {            dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);        }        cout << fixed << setprecision(2) << dis[0][1] << endl;    }    return 0;}

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