D - Area
Time Limit:1000MS
Memory Limit:10000KB
64bit IO Format:%I64d
& %I64u
Submit Status
Description
Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has installed the latest system of surveillance robots patrolling the
area. These robots move along the walls of the facility and report suspicious observations to the central security office. The only flaw in the system a competitor抯 agent could find is the fact that the robots radio their movements unencrypted. Not being able
to find out more, the agent wants to use that information to calculate the exact size of the area occupied by the new facility. It is public knowledge that all the corners of the building are situated on a rectangular grid and that only straight walls are
used. Figure 1 shows the course of a robot around an example area.
Figure 1: Example area.
You are hired to write a program that calculates the area occupied by the new facility from the movements of a robot along its walls. You can assume that this area is a polygon with cZ喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">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"pst">
Input
The first line contains the number of scenarios.
For each scenario, you are given the number m, 3 <= m < 100, of movements of the robot in the first line. The following m lines contain pairs 揹x dy?of integers, separated by a single blank, satisfying .-100 <= dx, dy <= 100 and (dx, dy) != (0, 0). Such a pair
means that the robot moves on to a grid point dx units to the right and dy units upwards on the grid (with respect to the current position). You can assume that the curve along which the robot moves is closed and that it does not intersect or even touch itself
except for the start and end points. The robot moves anti-clockwise around the building, so the area to be calculated lies to the left of the curve. It is known in advance that the whole polygon would fit into a square on the grid with a side length of 100
units.
Output
The output for every scenario begins with a line containing 揝cenario #i:? where i is the number of the scenario starting at 1. Then print a single line containing I, E, and A, the area A rounded to one digit after the decimal point. Separate the three numbers
by two single blanks. Terminate the output for the scenario with a blank line.
Sample Input
2
4
1 0
0 1
-1 0
0 -1
7
5 0
1 3
-2 2
-1 0
0 -3
-3 1
0 -3
Sample Output
Scenario #1:
0 4 1.0
Scenario #2:
12 16 19.0
這題有點……剛開始直接用了多邊形的模板，然後得不到樣例的答案，有點神了……而且我是照著計算幾何PDF上的代碼敲的，竟然不對，我還在想是不是坑我呢這代碼……
沒出樣例的原因是沒看好題目，題目給的是機器人移動的方位，而不是剛開始就給出坐標，所以它的坐標就是移動方位的累加咯，走到哪當然就是坐標咯。
#include
#include
#include
#include
#include
#define MAX 111116
#define eps 1e-7
using namespace std;
int sgn(const double &x){ return x < -eps? -1 : (x > eps);}
inline double sqr(const double &x){ return x * x;}
inline int gcd(int a, int b){ return !b? a: gcd(b, a % b);}
struct Point
{
double x, y;
Point(){}
Point(const double &x, const double &y):x(x), y(y){}
Point operator -(const Point &a)const{ return Point(x - a.x, y - a.y); }
Point operator +(const Point &a)const{ return Point(x + a.x, y + a.y); }
Point operator * (const double &a)const{ return Point(x * a, y * a); }
Point operator / (const double &a)const{ return Point(x / a, y / a); }
friend double det(const Point &a, const Point &b){ return a.x * b.y - a.y * b.x;}
friend double dot(const Point &a, const Point &b){ return a.x * b.x + a.y * b.y;}
friend double dist(const Point &a, const Point &b){ return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}
void in(){ scanf("%lf %lf", &x, &y); }
void out(){ printf("%.2f %.2f\n", x, y); }
};
struct Line
{
Point s, t;
Line() {}
Line(const Point &s, const Point &t):s(s), t(t) {}
void in() { s.in(),t.in(); }
double pointDistLine(const Point &p)
{
if(sgn(dot(t - s, p - s)) < 0)return dist(p, s);
if(sgn(dot( s - t, p - t)) < 0)return dist(p, t);
return fabs(det(t - s, p - s)) / dist(s, t);
}
bool pointOnLine(const Point &p)
{
return sgn(det(t - s, p - s)) == 0 && sgn(dot(s - p, t - p)) <= 0;
}
};
struct Poly //多邊形類
{
vectora;
void in(const int &r)
{
a.resize(r + 1);
for(int i = 1; i <= r; i++)
{
a[i].in(); //點坐標累加，這個剛開始不懂，所以樣例一直出不來
a[i] = a[i - 1] + a[i]; //因為是移動的方位，所以坐標就是累加的了
}
a.resize(r);
}
//計算多邊形的周長
double perimeter()
{
double sum=0;
int n=a.size();
for(int i=0;i0&&d1<=0&&d2>0) num++;
if(k<0&&d2<=0&&d1>0) num--;
}
return num!=0;
}
//計算多邊形邊界的格點數
int border()
{
int num=0,i,n=a.size();
for(i=0;i>T;
for(i=1;i<=T;i++)
{
int n;
cin>>n;
poly.in(n);
printf("Scenario #%d:\n",i);
printf("%d %d %.1f\n\n",poly.inside(),poly.border(),poly.getDArea());
}
return 0;
}
