题目： 传送门
题意：在一个左小角坐标为(0, 0)，右上角坐标为(10, 10)的房间里，有 n 堵墙，每堵墙都有两个门。每堵墙的输入方式为 x, y1, y2, y3, y4，x 是墙的横坐标，第一个门的区间为[ (x, y1) ~ (x, y2) ]，问你从 (0, 5) 走到 (10, 5) 的最短路径是多少。
0 <= n <= 18
题解：讲每个门的端点存起来，然后，加上(0, 5) 和 (10, 5) 这两个点，暴力判断这些点两两间是否能互相直达，然后再跑一遍最短路就行了。
你不能直接从 (0, 5) 到 （10, 5） 那你肯定是经过一些门的端点再到终点是最优的。
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <map> #include <vector> #include <set> #include <string> #include <math.h> #define LL long long #define mem(i, j) memset(i, j, sizeof(i)) #define rep(i, j, k) for(int i = j; i <= k; i++) #define dep(i, j, k) for(int i = k; i >= j; i--) #define pb push_back #define make make_pair #define INF 1e20 #define inf LLONG_MAX #define PI acos(-1) using namespace std; const int N = 1e2 + 5; const double eps = 1e-10; struct Point { double x, y; Point(double x = 0, double y = 0) : x(x), y(y) { } /// 构造函数 }; /// 向量加减乘除 inline Point operator + (const Point& A, const Point& B) { return Point(A.x + B.x, A.y + B.y); } inline Point operator - (const Point& A, const Point& B) { return Point(A.x - B.x, A.y - B.y); } inline Point operator * (const Point& A, const double& p) { return Point(A.x * p, A.y * p); } inline Point operator / (const Point& A, const double& p) { return Point(A.x / p, A.y / p); } inline int dcmp(const double& x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } inline double Cross(const Point& A, const Point& B) { return A.x * B.y - A.y * B.x; } /// 叉积 inline double Dot(const Point& A, const Point& B) { return A.x * B.x + A.y * B.y; } /// 点积 inline double Length(const Point& A) { return sqrt(Dot(A, A)); } /// 向量长度 inline double Angle(const Point& A, const Point& B) { return acos(Dot(A, B) / Length(A) / Length(B)); } /// 向量A,B夹角 inline Point GetLineIntersection(const Point P, const Point v, const Point Q, const Point w) {///求直线p + v*t 和 Q + w*t 的交点，需确保有交点，v和w是方向向量 Point u = P - Q; double t = Cross(w, u) / Cross(v, w); return P + v * t; } inline bool Onsegment(Point p, Point a1, Point a2) { /// 判断点p是否在线段p1p2上 return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) <= 0; } inline bool SegmentProperInsection(Point a1, Point a2, Point b1, Point b2) { /// 判断线段是否相交 if(dcmp(Cross(a1 - a2, b1 - b2)) == 0) /// 两线段平行 return Onsegment(b1, a1, a2) || Onsegment(b2, a1, a2) || Onsegment(a1, b1, b2) || Onsegment(a2, b1, b2); Point tmp = GetLineIntersection(a1, a2 - a1, b1, b2 - b1); return Onsegment(tmp, a1, a2) && Onsegment(tmp, b1, b2); } Point P[N]; double dis[N][N]; int main() { int n; double x, y1, y2, y3, y4; while(scanf("%d", &n) && n != -1) { int cnt = 0; rep(i, 1, n) { scanf("%lf %lf %lf %lf %lf", &x, &y1, &y2, &y3, &y4); P[++cnt] = Point(x, y1); P[++cnt] = Point(x, y2); P[++cnt] = Point(x, y3); P[++cnt] = Point(x, y4); } rep(i, 0, cnt + 1) rep(j, 0, cnt + 1) if(i == j) dis[i][j] = 0; else dis[i][j] = INF; rep(i, 1, cnt) { /// 判断(0,5)是否能直达P[i]，P[i]是否能直达(10,5) bool flag = 0; int up = ((i + 3) / 4) * 4 + 1; up -= 4; for(int j = 1; j < up; j += 4) { /// (0,5)是否能直达P[i] if(SegmentProperInsection(P[j], P[j + 1], Point(0, 5), P[i]) == false && SegmentProperInsection(P[j + 2], P[j + 3], Point(0, 5), P[i]) == false) flag = 1; } if(!flag) dis[0][i] = dis[i][0] = Length(Point(0, 5) - P[i]); flag = 0; for(int j = up + 4; j <= cnt; j += 4) { /// P[i] 是否能直达(10,5) if(SegmentProperInsection(P[j], P[j + 1], Point(10, 5), P[i]) == false && SegmentProperInsection(P[j + 2], P[j + 3], Point(10, 5), P[i]) == false) flag = 1; } if(!flag) dis[cnt + 1][i] = dis[i][cnt + 1] = Length(Point(10, 5) - P[i]); } rep(i, 1, cnt) rep(j, i + 1, cnt) { /// 枚举两点，判断这两点是否能直达 int st = ((i + 3) / 4) * 4 + 1; int ed = ((j + 3) / 4) * 4; bool flag = 0; for(int k = st; k <= ed; k += 4) { if(SegmentProperInsection(P[k], P[k + 1], P[i], P[j]) == false && SegmentProperInsection(P[k + 2], P[k + 3], P[i], P[j]) == false) flag = 1; } if(!flag) dis[i][j] = dis[j][i] = Length(P[i] - P[j]); } bool flag = 0; for(int i = 1; i <= cnt; i += 4) /// 判断(0,5)是否能直达(10,5) if(SegmentProperInsection(P[i], P[i + 1], Point(0, 5), Point(10, 5)) == false && SegmentProperInsection(P[i + 2], P[i + 3], Point(0, 5), Point(10, 5)) == false) flag = 1; if(!flag) dis[0][cnt + 1] = dis[cnt + 1][0] = 10; rep(k, 0, cnt + 1) rep(i, 0, cnt + 1) rep(j, 0, cnt + 1) if(dis[i][k] + dis[k][j] < dis[i][j]) dis[i][j] = dis[i][k] + dis[k][j]; printf("%.2f\n", dis[0][cnt + 1]); } return 0; }