POJ 1039 Pipe

枚举任意两个点作为直线。

看是否能穿越整个管道即可。

判断相交使用叉积。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <cmath>#include <map>#include <sstream>#include <queue>#include <vector>#define MAXN 100005#define MAXM 211111#define eps 1e-8#define INF 50000001using namespace std;inline int dblcmp(double d){    if(fabs(d) < eps) return 0;    return d > eps ? 1 : -1;}struct point{    double x, y;    point(){}    point(double _x, double _y): x(_x), y(_y) {}    void input()    {        scanf("%lf%lf", &x, &y);    }    bool operator ==(point a)const    {        return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;    }    point sub(point p)    {        return point(x - p.x, y - p.y);    }    double dot(point p)    {        return x * p.x + y * p.y;    }    double det(point p)    {        return x * p.y - y * p.x;    }    double distance(point p)    {        return hypot(x - p.x, y - p.y);    }}a[25], b[25];struct line{    point a, b;    line(){}    line(point _a, point _b){ a = _a; b = _b;}    void input()    {        a.input();        b.input();    }    int segcrossseg(line v)    {        int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));        int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));        int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));        int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));        if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2)return 2;        return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <= 0||                d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <= 0||                d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0||                d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);    }    int linecrossseg(line v)//v is seg    {        int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));        int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));        if ((d1 ^ d2) == -2) return 2;        return (d1 == 0 || d2 == 0);    }    point crosspoint(line v)    {        double a1 = v.b.sub(v.a).det(a.sub(v.a));        double a2 = v.b.sub(v.a).det(b.sub(v.a));        return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1));    }};int n;double ans;int gao(line x, int id){    int ind = -1;    int kind = -1;    for(int i = 0; i < n - 1; i ++)    {        if(dblcmp(x.b.sub(x.a).det(a[i].sub(x.a))) < 0 || dblcmp(x.b.sub(x.a).det(a[i + 1].sub(x.a))) < 0)        {            ind = i;            kind = 1;            break;        }        if(dblcmp(x.b.sub(x.a).det(b[i].sub(x.a))) > 0 || dblcmp(x.b.sub(x.a).det(b[i + 1].sub(x.a))) > 0)        {            ind = i;            kind = 2;            break;        }    }    if(ind != -1 && ind < id) return 0;    if(ind == -1) return 1;    line y;    if(kind == 1) y = line(a[ind], a[ind + 1]);    else y = line(b[ind], b[ind + 1]);    point c = x.crosspoint(y);    ans = max(ans, c.x);    return 0;}int main(){    while(scanf("%d", &n) != EOF && n)    {        for(int i = 0; i < n; i++)        {            a[i].input();            b[i].x = a[i].x;            b[i].y = a[i].y - 1;        }        ans = a[0].x;        int flag = 0;        for(int i = 0; i < n; i++)            for(int j = i + 1; j < n; j++)            {                line x = line(a[i], b[j]);                flag |= gao(x, j);                x = line(b[i], a[j]);                flag |= gao(x, j);                if(flag) break;            }        if(flag) puts("Through all the pipe.");        else printf("%.2f\n", ans);    }    return 0;}