poj 1265 pick定理的应用—Area
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Description
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
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
Sample Input
241 00 1-1 00 -175 01 3-2 2-1 00 -3-3 10 -3
Sample Output
Scenario #1:0 4 1.0Scenario #2:12 16 19.0
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int m,t,I,E;
double area;
struct Point
{
int x,y;
}P[105];
int Gcd(int a,int b)
{
return b?Gcd(b,a%b):a;
}
int corss(int x,int y,int x1,int y1)
{
return x1*y-x*y1;
}
int dx,dy;
int main()
{
int i,j,k;
scanf("%d",&t);
k=1;
while(t--)
{
scanf("%d",&m);
P[0].x=P[0].y=0;
area=0;E=0;
for(i=1;i<=m;i++)
{
scanf("%d%d",&dx,&dy);
P[i].x=dx+P[i-1].x;
P[i].y=dy+P[i-1].y;
E+=Gcd(abs(dx),abs(dy));
area+=corss(P[i].x,P[i].y,P[i-1].x,P[i-1].y);//面积的二倍
}
// area=b/2+i-1;2*area=b+2*i-2
I=(area+2-E)/2;
printf("Scenario #%d:\n%d %d %.1lf\n\n",k++,I,E,area/2.0);
}
return 0;
}
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