vjudge链接:https://vjudge.net/problem/POJ-2826
题意很简单,天上下雨,用两根线段接水,问能接住多少面积的水。
看起来很简单,然后WA了5发。。。
首先判断两根线段是否相交,若不相交则显然不能接到水。
然后求交点p。分别求出两条线段的端点中较上面的端点p1,
p2。然后以交点为新的原点,p1 -= p, p2 -=
p。这样就映射到了以p为原点的坐标系中。问题转化为p1、原点、p2形成的碗能不能接到水。
注意雨是从天下下的,因此需要特判一下较高的线段会不会把较低的线段挡住。
y = min(p1.y, p2.y),显然y>=0
如果y=0,那么显然不能接到水。
y>0时,安排p1和p2的顺序,使得p1.y <=
p2.y。然后看一下p1.x和p2.x是否在同一侧,如果不在则一定不会挡住。如果都在左侧,则看一下p1O是不是在p2O左侧以及fabs(p1.x)是否小于等于fabs(p2.x),若是,则说明p1被p2挡住了。右侧同理。
没有被挡住的情况就正常求了。不赘述。
赠送一波数据:
100
0 1 1 0
1 0 2 1
0 1 2 1
1 0 1 2
0 0 1 1
2 0 0 2
0 1 1 0
2 0 3 1
2 0 0 2
1 1 2 2
-10000 10000 0 -10000
0 -10000 10000 10000
0 0 3 3
0 2 2 0
0 3 3 0
1 0 3 2
0 3 3 0
1 1 3 0
0 0 2 1
0 0 3 3
0 0 3 3
0 0 3 3
0 2 2 0
0 1 2 0
0 0 2 1
0 0 2 2
0 0 3 3
0 0 1 2
>>
1.00
0.00
0.00
0.00
1.00
200000000.00
1.00
1.00
0.00
0.00
0.00
0.00
0.00
1.00
代码:
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
#define MAXN 100011
template <typename T>
(T x) {
T sqreturn x * x;
}
const double eps = 1e-8, pi = acos(-1.0);
int sgn(double x) {
return x < -eps ? -1 : (x > eps ? 1 : 0);
}
struct VEC {
double x, y;
bool operator == (const VEC& b) const {
return sgn(x - b.x) == 0 && sgn(y - b.y) == 0;
}
bool operator < (const VEC& b) const {
return sgn(y - b.y) == 0 ? sgn(x - b.x) < 0 : y < b.y;
}
operator - (const VEC& b) const {
VEC return VEC{x - b.x, y - b.y};
}
operator + (const VEC& b) const {
VEC return VEC{x + b.x, y + b.y};
}
double len() const {
return hypot(x, y);
}
double dist(const VEC& b) const {
return (*this - b).len();
}
double len2() const {
return sq(x) + sq(y);
}
double dist2(const VEC& b) const {
return (*this - b).len2();
}
operator * (double k) const {
VEC return VEC{x * k, y * k};
}
(double l) const {
VEC truncdouble ori = len();
if (!sgn(ori)) return *this;
return *this * (l / ori);
}
operator / (double k) const {
VEC return VEC{x / k, y / k};
}
() const {
VEC normreturn *this / len();
}
double operator ^ (const VEC& b) const {
return x * b.y - y * b.x;
}
double operator * (const VEC& b) const {
return x * b.x + y * b.y;
}
// The angle between *this and b
//anticlockwise is plus
double rad_di(const VEC& b) const {
return atan2(*this ^ b, *this * b);
}
double rad(const VEC& b) const {
return fabs(rad_di(b));
//return asin(fabs(*this ^ b) / (len() * b.len()));
}
() const {
VEC rotleftreturn VEC{-y, x};
}
() const {
VEC rotrightreturn VEC{y, -x};
}
(double a) const {
VEC rotdouble c = cos(a), s = sin(a);
return VEC{x * c - y * s, x * s + y * c};
}
//Independent
//0 <= slope < pi
double slope() const {
double ans = atan2(y, x);
if (sgn(ans) < 0) ans += pi;
else if (sgn(ans - pi) == 0) ans -= pi;
return ans;
}
double cross(const VEC& p1, const VEC& p2) const {
return (p1 - *this) ^ (p2 - *this);
}
//-1: left
//0: parallel
//1: right
int relation(const VEC& b) const {
return sgn(b ^ *this);
}
operator - () const {
VEC return VEC{-x, -y};
}
bool SameDi(const VEC& b) const {
return sgn(x) == sgn(b.x) && sgn(y) == sgn(b.y);
}
void input() {
("%lf%lf", &x, &y);
scanf}
};
struct LINE {
, v;
VEC s() {}
LINE(const VEC& _s, const VEC& _v) : s(_s), v(_v) {}
LINEvoid set(const VEC& _s, double a) {
= _s;
s = {cos(a), sin(a)};
v }
//ax + by + c = 0
void set(double a, double b, double c) {
= {b, -a};
v = fabs(a) > fabs(b) ? VEC{-c / a, 0} : VEC{0, -c / b};
s }
//-1: left
//0: on it
//1: right
int relation(const VEC& p) const {
return sgn((p - s) ^ v);
}
//0: parallel
//1: coincide
//2: intersect
int relation(const LINE& b) const {
if (v.relation(b.v) == 0)
return b.relation(s) == 0;
return 2;
}
//Assume that they intersect
(const LINE& b) const {
VEC CrossPointdouble t = ((b.s - s) ^ b.v) / (v ^ b.v);
return s + v * t;
}
//left is plus
double dist_di(const VEC& p) const {
return (v ^ (p - s)) / v.len();
}
double dist(const VEC& p) const {
return fabs(dist_di(p));
}
(const VEC& p) const {
VEC projection= v.norm();
VEC v1 return v1 * ((p - s) * v1) + s;
}
(const VEC& p) const {
VEC symmetry= projection(p);
VEC p0 return p0 * 2 - p;
}
void SetAngularBisector(const VEC& a, const VEC& b, const VEC& c) {
(a, (atan2(b.y - a.y, b.x - a.x) + atan2(c.y - a.y, c.x - a.x)) / 2);
set}
};
struct SEG : LINE {
;
VEC e() {}
SEG(const VEC& _s, const VEC& _e) : e(_e), LINE(_s, _e - _s) {}
SEG
bool exclusive_between(const VEC& p) const {
return sgn((p - s) * (p - e)) < 0;
}
bool between(const VEC& p) const {
return sgn((p - s) * (p - e)) <= 0;
}
bool OnSeg(const VEC& p) const {
return sgn((p - s) ^ v) == 0 && between(p);
}
//0: Not cross
//1: Just cross (unnorm cross 非规范相交)
//2: Cross inner (norm cross 规范相交)
int relation(const SEG& b) const {
int d1 = sgn(v ^ (b.s - s));
int d2 = sgn(v ^ (b.e - s));
int d3 = sgn(b.v ^ (s - b.s));
int d4 = sgn(b.v ^ (e - b.s));
if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
else return (d1 == 0 && between(b.s)) || (d2 == 0 && between(b.e)) ||
(d3 == 0 && b.between(s)) || (d4 == 0 && b.between(e));
}
int relation(const LINE& b) const {
int d1 = sgn(b.v ^ (s - b.s));
int d2 = sgn(b.v ^ (e - b.s));
if ((d1 ^ d2) == -2) return 2;
else return d1 == 0 || d2 == 0;
}
double dist(const VEC& p) const {
if (sgn((p - s) * v) < 0 || sgn((p - e) * v) > 0)
return min(p.dist(s), p.dist(e));
return LINE::dist(p);
}
double dist(const SEG& b) const {
return relation(b) ? 0 : min(min(dist(b.s), dist(b.e)), min(b.dist(s), b.dist(e)));
}
};
double GetX(const VEC& v, double y) {
return v.x / v.y * y;
}
int main() {
int T;
("%d", &T);
scanfwhile (T--) {
, b;
SEG a.s.input();
a.e.input();
a.v = a.e - a.s;
a.s.input();
b.e.input();
b.v = b.e - b.s;
bbool fail = false;
if (!a.relation(b)) {
= true;
fail } else {
= a.CrossPoint(b);
VEC p = a.s.y > a.e.y ? a.s : a.e;
VEC p1 = b.s.y > b.e.y ? b.s : b.e;
VEC p2 = p1 - p;
p1 = p2 - p;
p2 double y = min(p1.y, p2.y);
if (sgn(y)) {
if (p1.y > p2.y) {
(p1, p2);
swap}
int side = sgn(p1.x);
if (0 == (side ^ sgn(p2.x)) && p2.relation(p1) == side && sgn(fabs(p1.x) - fabs(p2.x)) <= 0) {
= true;
fail }
if (!fail) {
double x1 = GetX(p1, y), x2 = GetX(p2, y);
if (x1 > x2) swap(x1, x2);
("%.2f\n", (x2 - x1) * y / 2 + eps);
printf}
} else {
= true;
fail }
}
if (fail) {
("0.00");
puts}
}
return 0;
}