POJ1039 Pipe

嘟嘟嘟

大致题意：按顺序给出\(n\)个拐点表示一个管道，注意这些点是管道的上端点，下端点是对应的\((x_i, y_i - 1)\)。从管道口射进一束光，问能达到最远的位置的横坐标。若穿过管道，输出\(Through\) \(all\) \(the\) $ pipe.$

还是线段求交问题。

枚举端点作为直线(光束)上的两个点。然后判断这条直线和每一条线段\((x_i, y_i)(x_i, y_i - 1)\)是否有交点。若无，则求出最远能到达的\(x\)。

注意坐标可为负，所以刚开始的极小值为\(-INF\)，而不是\(0\)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(‘ ‘)
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 25;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ‘ ‘;
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar();
  if(last == ‘-‘) ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar(‘-‘);
  if(x >= 10) write(x / 10);
  putchar(x % 10 + ‘0‘);
}

int n;
struct Vec
{
  db x, y;
  db operator * (const Vec& oth)const
  {
    return x * oth.y - oth.x * y;
  }
};
struct Point
{
  db x, y;
  Vec operator - (const Point& oth)const
  {
    return (Vec){x - oth.x, y - oth.y};
  }
}a[maxn], b[maxn];

db calc(Point A, Point B, Point C, Point D)
{
  Vec AB = B - A, AC = C - A, AD = D - A, CD = D - C;
  db s1 = fabs(AB * AC), s2 = fabs(AB * AD);
  return C.x + CD.x / (s1 + s2) * s1;
}
db solve(Point A, Point B)
{
  Vec AB = B - A;
  for(int i = 1; i <= n; ++i)
    {
      Vec AC = a[i] - A, AD = b[i] - A;
      if((AC * AB) * (AD * AB) > eps)
    {
      if(i == 1) return -INF;
      Vec AE = a[i - 1] - A;
      if((AE * AB) * (AC * AB) < -eps) return calc(A, B, a[i - 1], a[i]);
      Vec AF = b[i - 1] - A;
      if((AF * AB) * (AD * AB) < -eps) return calc(A, B, b[i - 1], b[i]);
      return -INF;
    }
    }
  return INF;
}

int main()
{
  while(scanf("%d", &n) && n)
    {
      for(int i = 1; i <= n; ++i)
    scanf("%lf%lf", &a[i].x, &a[i].y), b[i].x = a[i].x, b[i].y = a[i].y - 1;
      db ans = -INF;
      for(int i = 1; i < n && ans != INF; ++i)
    {
      for(int j = i + 1; j <= n; ++j)
        {
          ans = max(ans, solve(a[i], a[j]));
          if(ans == INF) break;
          ans = max(ans, solve(a[i], b[j]));
          if(ans == INF) break;
          ans = max(ans, solve(b[i], a[j]));
          if(ans == INF) break;
          ans = max(ans, solve(b[i], b[j]));
          if(ans == INF) break;
        }
    }
      if(ans == INF) puts("Through all the pipe.");
      else printf("%.2f\n", ans);
    }
  return 0;
}

原文地址：https://www.cnblogs.com/mrclr/p/9977923.html