POJ 1039 Pipe 枚舉線段相交

Pipe

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9493		Accepted: 2877

Description

The GX Light Pipeline Company started to prepare bent pipes for the new transgalactic light pipeline. During the design phase of the new pipe shape the company ran into the problem of determining how far the light can reach inside each component of the pipe. Note that the material which the pipe is made from is not transparent and not light reflecting. 

Each pipe component consists of many straight pipes connected tightly together. For the programming purposes, the company developed the description of each component as a sequence of points [x1; y1], [x2; y2], . . ., [xn; yn], where x1 < x2 < . . . xn . These are the upper points of the pipe contour. The bottom points of the pipe contour consist of points with y-coordinate decreased by 1. To each upper point [xi; yi] there is a corresponding bottom point [xi; (yi)-1] (see picture above). The company wants to find, for each pipe component, the point with maximal x-coordinate that the light will reach. The light is emitted by a segment source with endpoints [x1; (y1)-1] and [x1; y1] (endpoints are emitting light too). Assume that the light is not bent at the pipe bent points and the bent points do not stop the light beam.

Input

The input file contains several blocks each describing one pipe component. Each block starts with the number of bent points 2 <= n <= 20 on separate line. Each of the next n lines contains a pair of real values xi, yi separated by space. The last block is denoted with n = 0.

Output

The output file contains lines corresponding to blocks in input file. To each block in the input file there is one line in the output file. Each such line contains either a real value, written with precision of two decimal places, or the message Through all the pipe.. The real value is the desired maximal x-coordinate of the point where the light can reach from the source for corresponding pipe component. If this value equals to xn, then the message Through all the pipe. will appear in the output file.

Sample Input

4
0 1
2 2
4 1
6 4
6
0 1
2 -0.6
5 -4.45
7 -5.57
12 -10.8
17 -16.55
0

Sample Output

4.67
Through all the pipe.

Source

Central Europe 1995

題目大意：給出一個管道，讓一束光穿過，找出一個穿過距離最長的，而後求出這個穿過距離最長的與管壁的交點的x座標，若是能夠完整的穿過整個管道，輸出Through all the pipe.

思路： 枚舉光線是否能夠穿過每個拐點官腔，也就是拐點所在位置的豎直線段，若是不能穿過說明它確定與管壁相交，計算交點，若是都能相交，這時就是能夠穿過整個管道。

注：剛開始枚舉的時候枚舉的特別麻煩，是直接枚舉光線方向，從入口一次加0.01的枚舉，判斷特殊位置的時候特別麻煩

/*************************************************************************
    > File Name:            poj_1039.cpp
    > Author:               Howe_Young
    > Mail:                 1013410795@qq.com
    > Created Time:         2015年05月01日 星期五 09時43分46秒
 ************************************************************************/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define EPS 1e-8
#define INF 1e6
using namespace std;
struct point{
    double x, y;
};
const int maxn = 100;
point p[maxn];
int n;
int sgn(double x)
{
    if (fabs(x) < EPS)
        return 0;
    return x < 0 ? -1 : 1;
}
double x_multi(point p1, point p2, point p3)
{
    return (p3.x - p1.x) * (p2.y - p1.y) - (p2.x - p1.x) * (p3.y - p1.y);
}
void get_intersection(point p1, point p2, point p3, point p4, double &x, double &y)
{
    double a1, b1, c1, a2, b2, c2;//求交點過程
    a1 = (p2.y - p1.y) * 1.0;
    b1 = (p1.x - p2.x) * 1.0;
    c1 = (p2.x * p1.y - p1.x * p2.y) * 1.0;
    a2 = (p4.y - p3.y) * 1.0;
    b2 = (p3.x - p4.x) * 1.0;
    c2 = (p3.y * p4.x - p4.y * p3.x) * 1.0;
    x = (b1 * c2 - b2 * c1) / (b2 * a1 - b1 * a2);
    y = (a1 * c2 - c1 * a2) / (a2 * b1 - a1 * b2);
}

bool check(point p1, point p2, point p3, point p4)//p1p2是否穿過豎着的p3p4，查看這條線是否與每個拐角處上下鏈接的線段都相交，包括端點
{
    double d1 = x_multi(p1, p2, p3);
    double d2 = x_multi(p1, p2, p4);
    return d1 * d2 <= 0;
}
bool check2(point p1, point p2, point p3, point p4)//同理看p3, p4這兩個點是否在p1p2兩側，端點不算
{
    double d1 = x_multi(p1, p2, p3);
    double d2 = x_multi(p1, p2, p4);
    return d1 * d2 < 0;
}
point does(point p1)//它的對應的下一個端點
{
    p1.y--;
    return p1;
}
int main()
{
    while (~scanf("%d", &n) && n)
    {
        for (int i = 0; i < n; i++)
        {
            scanf("%lf %lf", &p[i].x, &p[i].y);
        }
        point p0;
        double ans = p[0].x;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (i == j)
                    continue;
                if (check(p[i], does(p[j]), p[0], does(p[0])))//若是光線能夠從入口射進來
                {
                    for (int k = 1; k < n; k++)
                    {
                        if (!check(p[i], does(p[j]), p[k], does(p[k])))//若是走到k點這個拐點與管壁相交了，找出相交的點來
                        {
                            if (check2(p[i], does(p[j]), p[k], p[k - 1]))//若是與上壁相交
                            {
                                get_intersection(p[i], does(p[j]), p[k], p[k - 1], p0.x, p0.y);
                                if (ans < p0.x)
                                    ans = p0.x;
                                break;
                            }
                            if (check2(p[i], does(p[j]), does(p[k]), does(p[k - 1])))//若是與下壁相交
                            {
                                get_intersection(p[i], does(p[j]), does(p[k]), does(p[k - 1]), p0.x, p0.y);
                                if (ans < p0.x)
                                    ans = p0.x;
                                break;
                            }//若是都不相交的話，那麼說明是與上一段的端點相交
                            if (ans < p[k - 1].x)
                                ans = p[k - 1].x;
                            break;
                        }
                        if (k == n - 1)//若是走到最後都沒break，也就是相交，那麼說明能夠經過這個管道，直接讓他等於最後的x座標
                        {
                            ans = p[n - 1].x;
                        }
                    }
                }
            }
        }
        if (sgn(ans - p[n - 1].x) == 0)
        {
            puts("Through all the pipe.");
        }
        else
            printf("%.2f\n", ans);
    }
    return 0;
}

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