POJ1066 Treasure Hunt

　　原题传送：http://poj.org/problem?id=1066

　　不用管“门要在墙壁中央”这个条件，直接枚举每个分割区间的中点为进入点，然后再求该进入点与藏宝点的线段与所有线段的最少交点数（不管路线怎么折，这个贪心策略都是正确的）。注意n=0的情况，注意线段相交精度（去除在端点相交的情况）。

View Code

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <algorithm>
  4 #include <vector>
  5 #define eps 1e-10
  6 using namespace std;
  7  
  8 struct point
  9 {
 10     double x, y;
 11     point(){}
 12     point(double _x, double _y){x = _x; y = _y;}
 13 }seg[35][2];
 14  
 15 int n;
 16 double ex, ey;
 17 vector<double> vl, vr, vu, vd;  // 四个边缘墙壁
 18 
 19 double min(double a, double b){return a < b ? a : b;}
 20 double max(double a, double b){return a > b ? a : b;
 21 }
 22 bool inter(point a, point b, point c, point d)
 23 {
 24     if ( min(a.x, b.x) > max(c.x, d.x) ||
 25          min(a.y, b.y) > max(c.y, d.y) ||
 26          min(c.x, d.x) > max(a.x, b.x) ||
 27          min(c.y, d.y) > max(a.y, b.y) ) return 0;
 28     double h, i, j, k;
 29     h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
 30     i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
 31     j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
 32     k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
 33     return h * i < 0.0 && j * k < 0.0; // 相交点在端点的情况不能要
 34 }
 35  
 36 int cross(double x, double y)
 37 {
 38     int i, cnt;
 39     for(cnt = i = 0; i < n; i ++)
 40         if(inter(point(x, y), point(ex, ey), seg[i][0], seg[i][1]))
 41             cnt ++;
 42     return cnt;
 43 }
 44  
 45 int main()
 46 {
 47     int i, ans;
 48     scanf("%d", &n);
 49     for(i = 0; i < n; i ++)
 50     {
 51         scanf("%lf%lf%lf%lf", &seg[i][0].x, &seg[i][0].y, &seg[i][1].x, &seg[i][1].y);
 52         if(seg[i][0].x < eps)
 53             vl.push_back(seg[i][0].y);
 54         else if(seg[i][0].x - 100.0 < eps)
 55             vr.push_back(seg[i][0].y);
 56  
 57         if(seg[i][0].y < eps)
 58             vd.push_back(seg[i][0].x);
 59         else if(seg[i][0].x - 100.0 < eps)
 60             vu.push_back(seg[i][0].x);
 61  
 62         if(seg[i][1].x < eps)
 63             vl.push_back(seg[i][1].y);
 64         else if(seg[i][1].x - 100.0 < eps)
 65             vr.push_back(seg[i][1].y);
 66  
 67         if(seg[i][1].y < eps)
 68             vd.push_back(seg[i][1].x);
 69         else if(seg[i][1].x - 100.0 < eps)
 70             vu.push_back(seg[i][1].x);
 71     }
 72     scanf("%lf%lf", &ex, &ey);
 73     if(n == 0)
 74     {
 75         printf("Number of doors = 1\n");
 76         return 0;
 77     }
 78     sort(vl.begin(), vl.end());
 79     sort(vr.begin(), vr.end());
 80     sort(vd.begin(), vd.end());
 81     sort(vu.begin(), vu.end());
 82     ans = 100000000;
 83     for(i = 0; i < vl.size(); i ++)
 84     {
 85         int tmp = cross(0.0, i == 0 ? (vl[0] / 2) : (vl[i] + vl[i - 1]) / 2);
 86         ans = min(ans, tmp);
 87     }
 88     for(i = 0; i < vr.size(); i ++)
 89     {
 90         int tmp = cross(100.0, i == 0 ? (vr[0] / 2) : (vr[i] + vr[i - 1]) / 2);
 91         ans = min(ans, tmp);
 92     }
 93     for(i = 0; i < vu.size(); i ++)
 94     {
 95         int tmp = cross(i == 0 ? (vu[0] / 2) : (vu[i] + vu[i - 1]) / 2, 100.0);
 96         ans = min(ans, tmp);
 97     }
 98     for(i = 0; i < vd.size(); i ++)
 99     {
100         int tmp = cross(i == 0 ? (vd[0] / 2) : (vd[i] + vd[i - 1]) / 2, 0.0);
101         ans = min(ans, tmp);
102     }
103     printf("Number of doors = %d\n", ans + 1);
104     return 0;
105 }
106