poj1939(计算几何）

来源：互联网发布：外国人淘宝购物编辑：程序博客网时间：2024/06/05 19:58

http://poj.org/problem?id=1039

题意：在宽度为1的管里射条光线最远射多长？

思路：看了网上的题解，才明白叉积是能用来求相交的

#include <iostream>#include <cstdio>using namespace std;struct node{double x;double y;};node a[200];double max(double a, double b){return a>b?a:b;}node down(node p){p.y--;return p;}double xmult(node a, node b, node c){return (a.x - c.x)*(b.y - c.y) - (a.y-c.y)*(b.x-c.x);}bool check(node a, node b, node c, node d){double d1 = xmult(a, b, c);double d2 = xmult(a, b, d);if (d1*d2 <= 0) return true;return false;}bool checkl(node a, node b, node c, node d){double d1 = xmult(a, b, c);double d2 = xmult(a, b, d);if (d1*d2 >= 0) return false;return true;}node getPoint(node a, node b, node c, node d){node temp = a;double t = ((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));temp.x += (b.x-a.x)*t;temp.y += (b.y-a.y)*t;return temp;}int main(){int n, m, k;node t;double ans;while(cin >> n && n != 0){ans = 0;for (int i = 0; i < n; i++){cin >> a[i].x >> a[i].y; } double ans = a[0].x;for (int i = 0; i < n; i++) //upfor (int j = 0; j < n; j++){if (check(a[i], down(a[j]), a[0], down(a[0]))){for (k = 1; k < n; k++){if (!check(a[i], down(a[j]), a[k], down(a[k]))){if (checkl(a[i], down(a[j]), a[k], a[k-1])){t = getPoint(a[i], down(a[j]), a[k], a[k-1]);ans = max(t.x, ans);break;}if (checkl(a[i], down(a[j]), down(a[k]), down(a[k-1]))){t = getPoint(a[i], down(a[j]), down(a[k]), down(a[k-1]));ans = max(t.x, ans);break;}ans = max(ans, a[k-1].x);break;}}if (k == n){ans = a[n-1].x+1;}}}if (ans >= a[n-1].x){printf("Through all the pipe.\n");}else {printf("%.2f\n", ans);}}}

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