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POJ 2074 Line of Sight

2013-10-24 22:53 411 查看
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;
}
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