题目大意：从下面的一条直线看最上面的一条直线，中间有很多平行的障碍直线，问下面直线有多长是可以看到最上面的整条直线。。
思路：对于中间的每条直线求看不到的长度，最后求交集即可
code：
1 /* 2 State:Accepted 3 Time:2013-03-30 00:51:14 4 */ 5 #include <iostream> 6 #include <fstream> 7 #include <cstring> 8 #include <cstdio> 9 #include <cmath> 10 #include <algorithm> 11 #include <string> 12 #include <cstdlib> 13 #define eps 1e-8 14 using namespace std; 15 struct oo{double x1, x2, y;}; 16 17 double tx1 , ty ,tx2, dx1, dx2 ,dy; 18 oo a[1500], b[1500]; 19 int n; 20 void init(){ 21 scanf("%lf%lf%lf",&dx1, &dx2, &dy); 22 scanf("%d",&n); 23 for (int i = 1; i <= n ; ++i) 24 scanf("%lf%lf%lf",&a[i].x1, &a[i].x2, &a[i].y); 25 } 26 27 double work_x(double x1, double y1, double y2){ 28 double x2 = (x1*y2/y1); 29 return x2; 30 } 31 32 bool cmp(const oo a, const oo b){ 33 if (a.x1 < b.x1) return true; 34 if (fabs(a.x1 - b.x1) < eps && a.x2 > b.x2) return true; 35 return false; 36 } 37 38 void solve(){ 39 memset(b, 0 ,sizeof(b)); 40 for (int i = 1; i <= n ; ++i ){ 41 if (a[i].y >= ty - eps || a[i].y <= dy - eps) continue; 42 if ((a[i].y - dy) < eps){ 43 b[i].x1 = a[i].x1; 44 b[i].x2 = a[i].x2; 45 } else { 46 b[i].x2 = a[i].x2 - work_x(tx1 - a[i].x2, ty - a[i].y , a[i].y - dy); 47 b[i].x1 = a[i].x1 - work_x(tx2 - a[i].x1, ty - a[i].y , a[i].y - dy); 48 } 49 b[i].x2 = max(dx1,b[i].x2); 50 b[i].x2 = min(dx2,b[i].x2); 51 b[i].x1 = max(dx1,b[i].x1); 52 b[i].x1 = min(dx2, b[i].x1); 53 } 54 /*for (int i = 1; i <= n; ++i) 55 printf("%.2lf %.2lf\n", b[i].x1 ,b[i].x2);*/ 56 sort(b+1, b+n+1, cmp); 57 double ans =0, LL= 0 , x = dx1; 58 for (int i = 1; i <= n; ++i){ 59 if (b[i].x2 <= x) continue; 60 if (b[i].x1 > x){ 61 LL = b[i].x1 - x; 62 ans = max(ans , LL); 63 x = b[i].x1; 64 } 65 x = b[i].x2; 66 } 67 ans = max(ans , dx2 - x); 68 if (fabs(ans) < eps) printf("No View\n"); 69 else printf("%.2f\n",ans); 70 } 71 72 int main(){ 73 freopen("poj2074.in","r",stdin); 74 freopen("poj2074.out","w",stdout); 75 while (~scanf("%lf%lf%lf",&tx1, &tx2, &ty )){ 76 if (fabs(tx1)<eps && fabs(tx2)<eps && fabs(ty)<eps) break; 77 init(); 78 solve(); 79 } 80 fclose(stdin); fclose(stdout); 81 }