思路很清晰。
首先排除掉不在house和pro之间的障碍物
然后对每个障碍物，将站在直线上不能看到整个house的区间求出来。
最后对这些区间，扫一遍，贪心求一下即可
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <sstream>
#include <queue>
#include <vector>
#define MAXN 100005
#define MAXM 211111
#define eps 1e-8
#define INF 50000001
using namespace std;
inline int dblcmp(double d)
{if(fabs(d) < eps) return 0;return d > eps ? 1 : -1;
}
struct point
{double x, y;point(){}point(double _x, double _y): x(_x), y(_y) {}void input(){scanf("%lf%lf", &x, &y);}bool operator ==(point a)const{return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;}point sub(point p){return point(x - p.x, y - p.y);}double dot(point p){return x * p.x + y * p.y;}double det(point p){return x * p.y - y * p.x;}double distance(point p){return hypot(x - p.x, y - p.y);}
}p[5555];
struct line
{point a, b;line(){}line(point _a, point _b){ a = _a; b = _b;}void input(){a.input();b.input();}point crosspoint(line v){double a1 = v.b.sub(v.a).det(a.sub(v.a));double a2 = v.b.sub(v.a).det(b.sub(v.a));return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1));}}house, pro;
bool cmp(point a, point b)
{int k = dblcmp(a.x - b.x);if(k < 0) return 1;else if(k > 0) return 0;else return dblcmp(a.y - b.y) < 0;
}
int main()
{double xa, ya, xb;int m;while(scanf("%lf%lf%lf", &xa, &xb, &ya) != EOF){if(dblcmp(xa) == 0 && dblcmp(xb) == 0 && dblcmp(ya) == 0) break;house = line(point(xa, ya), point(xb, ya));scanf("%lf%lf%lf", &xa, &xb, &ya);pro = line(point(xa, ya), point(xb, ya));scanf("%d", &m);int cnt = 0;for(int i = 0; i < m; i++){scanf("%lf%lf%lf", &xa, &xb, &ya);if(dblcmp(ya - house.a.y) >= 0 || dblcmp(ya - pro.a.y) <= 0) continue;line tmp = line(point(xa, ya), house.b);point p1 = tmp.crosspoint(pro);tmp = line(point(xb, ya), house.a);point p2 = tmp.crosspoint(pro);if(dblcmp(p1.x - pro.b.x) > 0 || dblcmp(p2.x - pro.a.x) < 0) continue;p[cnt].x = max(pro.a.x, p1.x);p[cnt++].y = min(pro.b.x, p2.x);}double ans = 0;if(cnt){sort(p, p + cnt, cmp);double lst = pro.a.x;for(int i = 0; i < cnt; i++){if(dblcmp(p[i].x - lst) > 0) ans = max(ans, p[i].x - lst), lst = p[i].y;else if(dblcmp(p[i].y - lst) > 0) lst = p[i].y;}if(dblcmp(lst - pro.b.x) < 0) ans = max(ans, pro.b.x - lst);}else ans = pro.b.x - pro.a.x;if(dblcmp(ans) == 0) printf("No View\n");else printf("%.2f\n", ans);}return 0;
}