Area
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 5150 | Accepted: 2308 |
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 corners on a rectangular grid. However, your boss insists that you use a formula
he is so proud to have found somewhere. The formula relates the number I of grid points inside the polygon, the number E of grid points on the edges, and the total area A of the polygon. Unfortunately, you have lost the sheet on which he had written down that
simple formula for you, so your first task is to find the formula yourself.
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
Source
Northwestern Europe 2001
首先知识储备：1、匹克定理，在一个点阵中的多边形的面积s，和多边形内部的点的个数a，在多边形边上的点数b，存在关系s = a + b / 2 - 1 ;
2、向量的叉积a = (x1,y1) , b = (x2,y2).a×b = x1*y2 - y1*x2 ; 由矩阵得到。
3、求多边形面积，也就是求有向面积，使用叉积，求出有向的四边形的面积，除2得到三角形的面积，方法：找一个任意定点，开始遍历多边形，每过一条边，就计算叉积，得到对应三角形的面积，面积有正有负，也就对应着多边形的有向面积。即初始向量<a,b>存在一条边<b,c>求出<a,b>和<b,c>的叉积累加，然后更新初始向量为<a,c>。
有了这几个知识，这个题就好解决了，题目给了m个运动方向，即给出x，y代表从当前点运动(x,y)的距离。所以可以假设从（0,0）开始运动的。
在每一次运动中，得到的边上有多少个节点，可以通过gcd(x,y)得到，因为节点都是整数，所以x，y的最大公约数，也就是经过的节点数。然后求出总的多边形面积，最后得到多边形内部点数。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std ;
//匹克定理，s = a + b/2 - 1 ;
struct node{
int x , y ;
};
int gcd(int a,int b)
{
return b == 0 ? a : gcd(b,a%b) ;
}
node add(node p,node q)
{
p.x += q.x ;
p.y += q.y ;
return p;
}
double f(node p,node q)
{
double s = 0 ;
s = (p.x*q.y - p.y*q.x) / 2.0 ;
return s ;
}
int main()
{
int t , tt , m , a , b ;
double s ;
node p , q ;
scanf("%d", &t) ;
for(tt = 1 ; tt <= t ; tt++)
{
b = p.x = p.y = 0 ;
s = 0.0 ;
scanf("%d", &m) ;
scanf("%d %d", &q.x, &q.y) ;
b += gcd(abs(q.x),abs(q.y)) ;
p = add(p,q) ;
m-- ;
while( m-- )
{
scanf("%d %d", &q.x, &q.y) ;
b += gcd(abs(q.x),abs(q.y)) ;
s += f(p,q) ;
p = add(p,q) ;
}
printf("Scenario #%d:\n", tt) ;
printf("%d %d %.1lf\n", int(s+1-b/2.0) , b, s);
if( tt != t )
printf("\n") ;
}
return 0;
}