标签:== oss set 题目 clu include ret tar operator
题目:传送门
题意:有一个 n 个拐点的曲折的管道,你有一束光射进去(直射),问你最远能射到点的 x 坐标是多大。
1 <= n <= 20
思路:首先需要想到,这条线肯定是经过管道的一个上拐点和一个下拐点。
然后就枚举所有情况就行了。
#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 INT_MAX #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) { } }; int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } Point operator + (Point A, Point B) { return Point(A.x + B.x, A.y + B.y); } Point operator - (Point A, Point B) { return Point(A.x - B.x, A.y - B.y); } Point operator * (Point A, double p) { return Point(A.x * p, A.y * p); } Point operator / (Point A, double p) { return Point(A.x / p, A.y / p); } double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; } double Dot(Point A, Point B) { return A.x * B.x + A.y * B.y; } inline Point GLI(Point P, Point v, Point Q, 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 LSPI(Point b1, Point b2, Point a1, Point a2) { /// 判断直线b1b2是否与线段a1a3相交 return dcmp(Cross(b1 - a1, b2 - a1)) * dcmp(Cross(b1 - a2, b2 - a2)) <= 0; } Point up[N], dw[N]; int main() { int n; while(scanf("%d", &n) && n) { rep(i, 1, n) { scanf("%lf %lf", &up[i].x, &up[i].y); dw[i].x = up[i].x; dw[i].y = up[i].y - 1; } bool flag = 0; double ans = -100000000.0; rep(i, 1, n) rep(j, i + 1, n) { int pos; for(pos = 1; pos <= n; pos++) if(LSPI(up[i], dw[j], up[pos], dw[pos]) == false) break; if(pos > n) { flag = 1; break; } if(pos > j) { if(LSPI(up[i], dw[j], up[pos], up[pos - 1])) { Point tmp = GLI(up[i], dw[j] - up[i], up[pos], up[pos - 1] - up[pos]); ans = max(ans, tmp.x); } if(LSPI(up[i], dw[j], dw[pos], dw[pos - 1])) { Point tmp = GLI(up[i], dw[j] - up[i], dw[pos], dw[pos - 1] - dw[pos]); ans = max(ans, tmp.x); } } for(pos = 1; pos <= n; pos++) if(LSPI(dw[i], up[j], up[pos], dw[pos]) == false) break; if(pos > n) { flag = 1; break; } if(pos > j) { if(LSPI(dw[i], up[j], up[pos], up[pos - 1])) { Point tmp = GLI(dw[i], up[j] - dw[i], up[pos], up[pos - 1] - up[pos]); ans = max(ans, tmp.x); } if(LSPI(dw[i], up[j], dw[pos], dw[pos - 1])) { Point tmp = GLI(dw[i], up[j] - dw[i], dw[pos], dw[pos - 1] - dw[pos]); ans = max(ans, tmp.x); } } if(flag) break; } if(flag) puts("Through all the pipe."); else printf("%.2f\n", ans); } return 0; }
标签:== oss set 题目 clu include ret tar operator
原文地址:https://www.cnblogs.com/Willems/p/12392085.html