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题意：一根管子，中间有一些拐点，给出拐点的上坐标，下坐标为上坐标的纵坐标减1，管子不能透过光线也不能折射光线，问光线能射到最远的点的横坐标。
解法：光线射到最远处的时候一定最少经过两个拐点，枚举每两个顶点，判断最远光线射到的位置。代码姿势不够优美……都是眼泪啊
代码：
#include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<math.h> #include<limits.h> #include<time.h> #include<stdlib.h> #include<map> #include<queue> #include<set> #include<stack> #include<vector> #define LL long long using namespace std; const double eps = 1e-8, inf = 9999999.0; struct point { double x, y; } up[25], down[25]; int n; double cross(point p1, point p2, point p3) { return (p2.x - p1.x) * (p3.y - p1.y) - (p3.x - p1.x) * (p2.y - p1.y); } int judge(point a, point b) { for(int i = 0; i < n; i++) { double tmp = cross(a, b, up[i]) * cross(a, b, down[i]); if(tmp > eps) return i; } return -1; } int main() { while(~scanf("%d", &n) && n) { double ans = -inf; int flag = 0; for(int i = 0; i < n; i++) { double a, b; scanf("%lf%lf", &a, &b); up[i].x = down[i].x = a; up[i].y = b, down[i].y = b - 1.0; } for(int i = 0; i < n; i++) for(int j = i + 1; j < n; j++) { int tmp = judge(up[i], down[j]); if(tmp >= 0) { double res; if(tmp < j) continue; if(cross(up[i], down[j], up[tmp]) > 0) { double tmp1, tmp2; tmp1 = cross(up[i], down[j], down[tmp - 1]); tmp2 = cross(up[i], down[j], down[tmp]); res = (tmp2 * down[tmp - 1].x - tmp1 * down[tmp].x) / (tmp2 - tmp1); } else { double tmp1, tmp2; tmp1 = cross(up[i], down[j], up[tmp - 1]); tmp2 = cross(up[i], down[j], up[tmp]); res = (tmp2 * up[tmp - 1].x - tmp1 * up[tmp].x) / (tmp2 - tmp1); } ans = max(ans, res); } else flag = 1; } for(int i = 0; i < n; i++) for(int j = i + 1; j < n; j++) { int tmp = judge(down[i], up[j]); if(tmp >= 0) { double res; if(tmp < j) continue; if(cross(down[i], up[j], up[tmp]) > 0) { double tmp1, tmp2; tmp1 = cross(down[i], up[j], down[tmp - 1]); tmp2 = cross(down[i], up[j], down[tmp]); res = (tmp2 * down[tmp - 1].x - tmp1 * down[tmp].x) / (tmp2 - tmp1); } else { double tmp1, tmp2; tmp1 = cross(down[i], up[j], up[tmp - 1]); tmp2 = cross(down[i], up[j], up[tmp]); res = (tmp2 * up[tmp - 1].x - tmp1 * up[tmp].x) / (tmp2 - tmp1); } ans = max(ans, res); } else flag = 1; } if(flag) puts("Through all the pipe."); else printf("%.2f\n", ans); } return 0; }
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原文地址：http://www.cnblogs.com/Apro/p/4313839.html
