POJ 1039 （求线段交点 + 思维）

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;
}

原文地址：https://www.cnblogs.com/Willems/p/12392085.html